You can put this solution on YOUR website!

First let's put the equation into standard form. We'll start by simplifying the left side:

Then we'll gather all the terms on one side (so the other side is zero):

Next we will gather and group the terms, the x terms and the "other terms":

Factoring out from the first group and the x from the second group we get:

To make this look more like standard form I will use the Commutative Property to switch the order of the factors:

We now have standard form with...
a = m-1
b = m+1
c = m-1

We will get equal roots if (the discriminant) = 0. Replacing the a, b and c we found above into this equation we get:

Simplifying we get:




Now we solve for m. It will be easier to factor if we make the "a" in this quadratic positive. So we'll start by factoring out -1:

Then we can factor more:

From the Zero Product Property:
-1 = 0 or 3m-1 = 0 or m-3 = 0
The first equation is false and has no solution. The other two equations have solutions:
m = 1/3 or m = 3
So if the m in your original equation is either 1/3 or 3 there will be two equal roots to the equation.

To find the roots when m = 1/3:

Replace the m with 1/3:

Now we solve for x. To make things easier I'm going to multiply each side by three to get rid of the fraction:

Simplifying...

Making one side zero:

Factoring:


Zero Product Property:
2 = 0 or
There is no solution to the first equation. But the second equation has a solution of x = 1. So when m = 1/3 your equation has two equal roots of 1.

I'll leave it up to you to figure out the equal roots when m = 3.

You can put this solution on YOUR website!

Find find the oblique asymptote of a rational function you divide the numerator by the denominator so you can rewrite the function in the form:
f(x) = quotient + remainder/(f(x)'s denominator)

We will not be able to use synthetic division since our function's denominator is quadratic. So we must use long division. [Note: Since -6x^2 does not divide evenly into 7x^3 (and probably other terms too) this is going to get messy with fractions.]

              (-7/6)x + (-65/36)                  
-6x^2 -5x -9 / 7x^3       -5x^2            -9x + 7
             - 7x^3 + (35/6)x^2    +   (42/6)x
                        (-65/6)x^2 + (-294/6)x + 7
                      - (-65/6)x^2 + (325/36)x + 65/4
                                   (-2089/36)x + (-37/4)

So f(x) in the desired form is:

As x get to be very large (positive or negative), the fraction at the end becomes closer and closer to zero. This makes the quotient part the oblique asymptote for large values of x. So the oblique asymptote for f(x) is the line:

You can put this solution on YOUR website!
4x^9 + 8x^8 - 12x^7 + 8x^6
If you have trouble finding GCF's, it can be helpful to factor each term "fully". By "fully" I mean:

• Factor each coefficient into prime numbers
• Rewrite the variables without exponents

Let's see:

4x^9  = 2 * 2         * x * x * x * x * x * x * x * x * x
8x^8  = 2 * 2 * 2     * x * x * x * x * x * x * x * x
12x^7 = 2 * 2     * 3 * x * x * x * x * x * x * x
8x^6  = 2 * 2 * 2     * x * x * x * x * x * x

Note how I used spacing the ensure that each column has the same factor in it. Lining up the factors like this can also be helpful in finding the GCF. In the table above a factor that is common to all four terms will be the factor in any column that has four entries. The GCF will be the product of all the common factors. Looking at the table we can see that there are 2 columns of 2's and 6 columns of x's that have an entry in all four rows. So the GCF is:

GCF   = 2 * 2         * x * x * x * x * x * x = 4x^6

P.S. There are other uses for factoring and lining up the factors like this:

• Figuring out "what's left" of each term if the GCF is factored out. Just look at the GCF line of factors and compare it to a term's line. The factors that are in the term's line but not in the GCF's line of factors will be "what's left" of the term after the GCF is factored out. For example:
  

  GCF   = 2 * 2         * x * x * x * x * x * x
  8x^6  = 2 * 2 * 2     * x * x * x * x * x * x

  From this we can see that the only factor in the term's line that is not in the GCF line is that 3rd 2. So if the GCF is factored out of 8x^6 there will be a 2 left. Or
  

  GCF   = 2 * 2         * x * x * x * x * x * x
  12x^7 = 2 * 2     * 3 * x * x * x * x * x * x * x

  Here we can see that the term has factors of a 3 and a 7th x that are not in the GCF list. So if we factor out the GCF from this term we will have 3 * x or just 3x left.
  
  Note: If a term is the same as the GCF then there will be just a 1 (one) left if the GCF is factored out of this term.
  
  
• Finding least common multiples (LCM's). The LCM will be the product of all the different columns. For our four terms above the LCM will be:
  LCM = 2 * 2 * 2 * 3 * x * x * x * x * x * x * x * x * x = 24x^9
  Note: Lowest common denominators (LCD's) are simply the LCM of some denominators. So we can use this method to find LCD's, too. Not only that we can figure out each fraction need to be multiplied by to turn the denominator into the LCD. For example, let's say we wanted to add
  
  We would need the LCD. Since the denominators are the same as the terms we've already factored we can use the LCM we found above:
  

  LCM   = 2 * 2 * 2 * 3 * x * x * x * x * x * x * x * x * x = 12x^9

  Then if we compare the factors of the LCM with the factors of each denominator we can see what factors are "missing". For the first denominator:
  

  4x^9  = 2 * 2         * x * x * x * x * x * x * x * x * x

  We can see that this term's factors are "missing" the third 2 and a 3. So we would multiply the numerator and denominator of the first fraction by 2*3 or 6. For the third denominator:
  

  12x^7 = 2 * 2     * 3 * x * x * x * x * x * x * x

  We can see that this term's factors are missing the third 2 and the 8th and 9th x's. So we would multiply the third fraction's numerator and denominator by 2 * x * x or
  Note: If a denominator is the same as the LCD then it does not need to change. Don't multiply it by anything.

You can put this solution on YOUR website!

This is a compound fraction because of the negative exponents. Let's rewrite the expression with positive exponents so we can see it better:


Different methods are taught on how to simplify this expression. One method is to

1. Combine the terms in both the numerator and the denominator of the "big" fraction
2. Turn the division into a multiplication by the reciprocal of the denominator
3. Simplify

But I prefer a different method:

1. Find the lowest common denominator (LCD) of all the "little" fractions.
2. Multiply the numerator and denominator of the "big" fraction by the LCD from step one.

The first method requires addition/subtraction of fractions. The second one does not which is a major reason I prefer it.

Using the second method...
The "little" denominators are , , x and y. The LCD of these 4 is: . Multiplying the numerator and denominator by this LCD:

To multiply correctly we need to use the Distributive Property:

In each "little" fraction the denominator cancels with all or some part of the LCD leaving:

There are no like terms here so we cannot simplify this further. But we might be able to reduce the fraction. To see if the fraction will reduce we must first factor the numerator and denominator:

There are no factors in common between the numerator and denominator so the fraction will not reduce.

The problem asks that the answer be left in factored form so our answer is:

You can put this solution on YOUR website!
I think the reason no tutor has responded to this earlier is that they do not understand what you posted. I'm making a guess that the left side is:

If I am right then please try to make your expressions clearer. Different kinds of roots can be difficult to post. For these either:

• Use some English and parentheses to make it clear. For example, you could have used:
  81 * (5th root(27))
• Or teach yourself the syntax for Algebra.com's expression-drawing software. Click on the "Show source" link above and you can see what I typed to get the root to display nicely. (It starts the three left braces, "{", then the special syntax, then it ends with three right braces, "}".

If I am wrong, then stop reading and re-post your problem in a clearer way.

Solving equations where the unknown/variable is in an exponent is easiest if it is possible to rewrite the equation so that both sides are power of the same number. So we will look to see if we can rewrite both sides of our equation this way. Looking at the numbers involved, 81, 27 and 3, it should not take a long time to figure out that all three of them are powers of 3. ( and ) So we will be able to rewrite the equation in terms of powers of 3:

We can replace the radical with a fractional exponent:

Next we simplify. First the power of a power on the left side. The rule is to multiply the exponents:

Now we can multiply on the left side. The rule is to add the exponents:



We now have the equation in the desired form, two powers of 3 that are equal. The only way these powers of 3 can be equal is if the exponents are equal. So:

And we are finished.


P.S. In response to the question in your thank you note:

Both 1/7 and 49 are powers of 7:

Changing the radical to a fractional exponent:

Using the rules for exponents to simplify:

You can put this solution on YOUR website!
There are many ways to solve a system like this, including:

• Substitution Method
• Linear Combination (aka Elimination) Method
• Matrix methods, including:
  • Gaussian Elimination
  • Inverse Matrices
• Cramer's Rule (determinants)

Since it would not be helpful for me (or any tutor) to use a method you know nothing about, I think it would be best for you to re-post your question and include either:

• The method you would like to see; or
• A list of methods you have learned so the tutors can pick one you can recognize.

You can put this solution on YOUR website!


Usually the first thing with the elimination method is get opposites lined up. With the fractions and decimals, however, this will not be simple. So the first thing we will do is eliminate the fractions and decimals.

Multiplying both sides of the first equation by 10 will eliminate the decimals. And multiplying both sides of the second equation by 14 will eliminate the fractions:


which simplify to:



Now that the fractions and decimals are gone we can more easily create lined-up opposites. Multiplying the first equation by -3 will create lined-up opposite y-terms:


which simplifies to:



After the opposites are lined up the next step is to add the two equations together:

With the y terms gone we can solve for x:


Now that we have x we can find y. Usually I recommend using one of the original equations at this point. It is the safest way to find the second variable. (Using an equation you created might be a problem if you've made a mistake somewhere along the way.) But with the x being a fraction and the original equations having decimals or fractions with different denominators, I'm going to be lazy and take a chance using of the later equations (without fractions or decimals). I'm going to use:

Substituting in our value for x:

which simplifies to:

Multiplying by 13 to eliminate the fraction:

Adding 33:

Dividing by -13:


So the solution to your system is the point:
(, )

You can put this solution on YOUR website!
Pretend for a moment that the function was simply g(x) = -x then I hope you would recognize this as the equation for a line. A line in slope-intercept form. A line with a slope of -1 and a y-intercept of 0. One of the points on this pretend line is (3, -3)

Since the real g(x) is -x for all x's except 3, graph the line described above except for the point (3, -3)!. In other words. draw an small open circle (a small ring) at the point (3, -3). (This open circle indicates that this point is not included.) Then draw the rest of the line as normal. When you are finished it should look like the the line y = -x except that it has a "hole" in it at (3, -3).

To finish the graph just plot the point (3, 2). (This is a regular point, a dot, not the open circle we used at (3, -3).)

Your final graph should be a line with a hole in it at (3, -3) plus the extra point (3, 2) (which is nowhere near the line).

You can put this solution on YOUR website!
When you are not sure how to prove a Trig identity (like this), it can be helpful to convert any tan's, cot's, sec's and/or csc's into equivalent expressions in terms of of sin and/or cos.

With tan = sin/cos and sec = 1/cos your equation becomes:

We can simplify the fractions within a fraction on the left side bu multiplying the numerator and denominator of the "big" fraction by :

The cos's cancel:

leaving:

which simplifies to:

And we're finished!

You can put this solution on YOUR website!
I think the other tutor did not notice that three cards are to be drawn. His/her solution is only correct for drawing a single card.

To find the probability of all three cards being less than 5:

1. Find the individual probabilities of each card being less than 5; then
2. Multiply these probabilities.

The probability of the first card being less than 5 is, as the other tutor said, 4/10.

The probability of the second card being less than 5 is 3/9 since there are only three cards left that are less than 4 and there are only 9 cards altogether after the first card.

The probability of the third card being less than 5 is 2/8 since there are only two cards left that are less than 4 and there are only 8 cards altogether after the first two cards.

The probability of all three cards being less than 5 is the product of the three individual probabilities:

which simplifies down to:

You can put this solution on YOUR website!
The solution from the other tutor is correct. But it is only correct because the each side of the wall can be divided evenly by a corresponding side of a brick. So a full, proper solution will first determine this divisibility.

The 20cm side of the brick will divide evenly into the 4m side of the wall: 4m = 400cm and 400/20 = 20

The 12cm side of the brick will divide evenly into the 12m side of the wall: 12m = 1200cm and 1200/12 = 100

The 9cm side of the brick will divide evenly into the 72cm side of the wall: 72/9 = 8

This makes the number of bricks needed: 20*100*8 = 16000 (which matches the answer from the other tutor).

P.S. When checking for the divisibility, try all possible combinations. In this problem, the 20cm side of the brick divides into both the 12m and 4m sides of the wall. The 12cm side of the brick divides into both the 72cm and 12m sides of the wall. But the 9cm side of the brick only divides evenly into the 72cm side of the wall. This leaves the 12m for the 12cm and then the 20cm into the 4m.

P.P.S. If you cannot find this divisibility then it will not be possible to build the wall unless you break off parts of the bricks. This is a more difficult problem to solve and the solution from the other tutor, to divide the volumes, will not work correctly.

You can put this solution on YOUR website!
When a polynomial (this is not a rational function by the way) is written in standard form with the terms listed from highest degree to lowest (like yours), the possible rational roots are all the ratios, positive and negative, that can be formed using a factor of the constant term (at the end) in the numerator over a factor of the leading coefficient in the denominator.

Your constant term is 33. Its factors are 1, 3, 11 and 33.

Your leading coefficient is 9. Its factors are 1, 3 and 9.

The possible rational roots of this polynomial are all the possible ratios, positive and negative, we can form using a factor of 33 (1, 3, 11 or 33) on top over a factor of 9 (1, 3 or 9):
+1/1, +3/1, +11/1, +33/1,
+1/3, +3/3, +11/3, +33/3,
+1/9, +3/9, +11/9, +33/9
which reduce to:
+1, +3, +11, +33,
+1/3, +1, +11/3, +11,
+1/9, +1/3, +11/9, +11/3
Removing the duplicates:
+1, +3, +11, +33, +1/3, +11/3, +1/9, +11/9

So there are 16 possible rational roots to f(x).

You can put this solution on YOUR website!

To find the zeroes of an expression of degree 3 or more (like this one) we need to factor it.

When factoring, check the greatest common factor (GCF) first. The GCF here is 1 (which we rarely factor out).

After the GCF, we look to use any and all of the other factoring techniques. One of these techniques is factoring by the use of factoring patterns. And one of these patterns (the only one in fact) that has two terms with a "+" between them is:

As the pattern shows, we need a sum of two cubes to use this pattern. Is a sum of cubes? It's definitely a sum. Are and perfect cubes? With a little investigation we should find that and . So we can use the pattern with the "a" being "2x" and the "b" being "5". So , according to the patterns factors into:

(Note: My use of parentheses may seem excessive. But it is actually a good idea to use parentheses like this when making substitutions like we have done here. It helps us avoid easily-made mistakes.)
This simplifies to:

Neither factor will factor further no matter which techniques we try.

From the Zero Product Property we know that this product will be zero only be zero if one of the factors is zero. And if this product is zero then the expression from which we got these factors, will also be zero. So the x values that make these factors zero will be the zeroes of . Setting each factor to zero and then solving will lead to our solution. For the first factor:

Solving this we should get x = -5/2.

For the second factor:

We must use the Quadratic Formula:

simplifying...



At this point we can stop. The negative in the square root means that the solutions will be complex numbers. But the problem specifically asks for real solutions. So the only real zero to is -5/2.

P.S. In case you are curious about the complex zeroes I will go ahead and finish with the Quadratic Formula. Continuing the simplifying:






which is short for:
or
In standard "a + bi" form these would be:
or
So there are three zeroes: two complex (above) and the one real we found earlier (-5/2).

You can put this solution on YOUR website!

First of all, when there are radicals in a function (like this one) rewrite them using a fractional exponent instead of the radical before trying to find the function's derivative. (We do this because I've never heard of derivatives of radicals being taught. But we certainly learn how to find derivatives of a power.) So the first thing we do is rewrite the radical:


This is a chain rule problem. At the outermost level we have something raised to the 1/4th power. Inside that we have the sec of something. Inside that we have 3 times theta. (Note: I'm using the very vague term "something" on purpose (as I hope you will understand why as I go through the solution).

We start on the outside and work our way in. The outer most function is something to the 1/4th power. The derivative of this, according to the chain rule, is:
(1/4)*(something #1)^((1/4)-1)*(derivative of something #1)

"Something #1" is: sec(something #2). The "derivative of something #1", according to the chain rule, will then be:
sec(something #2)*tan(something #2)*(derivative of something #2)

"Something #2" is . The "derivative of something #2" is simply 3.

Putting all this together we get:
r' =
Simplifying...
The 3 at the end and the 1/4 in front make 3/4 and the 1/4 - 1 make -3/4:
r' =
Since the sec's have the same argument, , we can multiply them. The rule is to add exponents. The (unseen) exponent on the second sec is 1. Adding -3/4 and 1:
r' =
This may be acceptable as an answer. Or perhaps reverting to radical form would be preferred:
r' =
or, to ensure there is no confusion about whether the tan is inside or outside the radical:
r' =

You can put this solution on YOUR website!
If you posted the actual question you have then it is a bad question for several reasons:

• What is "root"? Square root? cube root? 4th root? etc.
• What is inside the radical (or whatever kind of root it is)? Just the 2? or 2/3?
• The inverse functions take a ratio as input and provide an angle as output. Depending on what "root 2 over 3" means, it may be a valid ratio to be input to the inverse sin. But inverse sin will output an angle which is not valid input to inverse cos.

If what you posted is not literally the problem given to you (which is waht I suspect), then please:

• Be more careful in posting. There is virtually no chance an inverse sin is input to an inverse cos.
• Don't just say "root". Tell us what kind of root.
• Use parentheses to tell us what is inside the radical of the root. Use
  square root(2)/3 (or just sqrt(2)/3) for
  and use
  square root(2/3) (or just sqrt(2/3)) for

You can put this solution on YOUR website!
This problem is just like your plane and ships problem. (You can even reuse the diagram as long as you change the angles and the heights!) So you can solve it the same way as I showed you in that problem.

The only substantial difference between this problem and the other one is that the angles are not whole numbers. To handle angles measured in degrees and minutes you must convert them into a decimal number of degrees.

Since there are 60 minutes in a degree, 34 degrees and 16 minutes is the same as degrees. Now we convert the fraction into a decimal by dividing 16/60. So 34 degrees 16 minutes, as a decimal number of degrees, is: 34.266666... Round this off as you see fit.

I'll leave it up to you to convert 54 degrees 12 minutes to a decimal number of degrees.

Then as you solve the problem you will be using these decimal numbers in your tan's.

You can put this solution on YOUR website!
For the diagram...

1. Draw two parallel horizontal lines. The top one will represent the path of the plane. The bottom one represents the surface of the water.
2. Pick a point on the top line (closer to the end of the line than the middle). This will represent the plane. Let's label it "P".
3. Pick two separate points on the bottom line (near the other end of the line from where P is). Make sure there is space between the two points. These points will represent the ships. Label the ship closer to P as "X" and the ship farther away as "Y".
4. Draw four line segments:
   • From P to X
   • From P to Y
   • From X straight up to the upper line. Label the upper endpoint of this segment as "Q".
   • From Y straight up to the upper line. Label the upper endpoint of this segment as "R".
Since the two lines are horizontal and segments XQ and YR are vertical, XQ and YR are perpendicular to the two horizontal lines. So draw the "box" indicator of a right angle at each point where XQ and YR intersect the horizontal lines. (IOW at X, Y, Q and R).
5. Since the lines are horizontal lines are parallel and since distance between lines is measured perpendicularly and since XQ and YR are perpendicular, both XQ and YR are 2000m in length. Write 2000 next to XQ and YR to indicate their length.
6. The angles of depression (which are always angles from a horizontal) are angles RPY and QPX. If you've been following these directions, angle RPY is the smaller angle. So label it 60 degrees. Angle QPX is the 70 degree angle. But since angle RPY is inside angle QPX it will not be easy to label angle QPX. Perhaps it would be better just to make a note outside the diagram that the measure of angle QPX is 70 degrees.

Our diagram is now complete. It should have two right triangles, RPY and QPX, and a rectangle, XYRQ. There is another triangle, PXY. But it is not a right triangle and we only know one part of this triangle. (Angle XPY is the difference between angle RPY and QPX so it is 10 degrees.) We need three parts, including at least one side, of a non-right triangle to find all the other parts. So if possible we will be using just the two right triangles and the rectangle to solve the problem.

The problem asks us to find the distance between the ships. In our diagram this should be the length of XY. XY is not a side of either of the right triangles so we will not be able to solve for it directly. We will have to find other lengths first and then use them to find XY.

XY is a side of the rectangle. So its length is the same as the opposite side, QR. If we can find QR then we have also found XY.

QR is the difference between PR and PQ. Since PR and PQ are sides of our right triangles and since we know the angles of depression and the vertical sides of these triangles we can use Trig to find PR and PQ.
In triangle RPY the vertical side, YR, is opposite to the the angle of depression, angle RPY. And the side we want to find is the adjacent side to the angle of depression. The Trig functions that involve a ratio of opposite and adjacent are tan and cot. Since we do not have a button on our calculator for cot, we will use tan:
(Note: PR is not P times R. It is a single number/variable that is the length of side PR.)
Solving for PR. Multiplying both sides by PR we get:

Dividing both sides by tan(60):

60 is a special angle so we could find the tan exactly without a calculator. (It's .) But we will use a calculator and get a decimal anyway since

• PR is not the answer to our problem; and
• we will be subtracting PQ from PR; and
• we do not have a special angle in triangle QPX so we will have to use our calculator to find PQ; and
• there's not much point in using an exact value (with a square root) when you have to subtract a decimal approximation from it to find your answer.

Replacing tan(60) with its decimal approximation:
(Note: Feel free to use more decimal places.)
Dividing:
PR = 1154.7

We will repeat this for PQ. The only difference is that the angle is 70, not 60:




PQ = 727.9

QR, as noted earlier, is the difference between PR and PQ:
QR = PR - PQ
QR = 1154.7 - 727.9
QR = 426.8

And since QR and XY are opposite sides in a rectangle, XY must be 426.8, too. So the distance between the ships is approximately 426.8 meters. (Note: You might get a number slightly different from this is you used more (or fewer) decimal places for your tan's.)

You can put this solution on YOUR website!
Since (f)x = –2x + 4 and g(x) = –6x – 7 and you want to find f(x) - g(x), just replace f(x) with -2x+4 and g(x) with -6x-7:
f(x) - g(x)
(-2x+4)-(-6x-7)
[Important! Notice the use of parentheses! It is an extremely good habit to use parentheses with making substitutions (especially if you are substituting in multiple terms like we are here).]

Since subtractions cause problems (they are harder than additions and they are not commutative so you can't change the order), I am going to change the subtractions to additions of the opposite:
(-2x+4)-(-6x+(-7))
(-2x+4)+(6x+7) (Note how the signs changed for both terms in the parentheses that had the "-" in front!)
Since we now have all addition we can freely change the order and grouping of the terms. Reordering and grouping the like terms together we get:
(-2x+6x)+(4+7)
Adding the like terms:
4x+11

So f(x) - g(x) = 4x+11

You can put this solution on YOUR website!

If this were

would it be easier? You probably know that this would be solved using two equations:
or

Well, the problem we have can be solved in exactly the same way!
or
The trick part is that we need the negative of the whole right side in the second equation. When the right side is just a number, like 35, you can just stick a "-" in front. But with multiple terms, like , we need to put parentheses around the right side. With the parentheses, the "-" in front will negate the entire expression, not just the .

We now have two quadratic equations to solve. Let's do them one at a time. With the first equation we want one side to be zero. Subtracting the entire left side from both sides we get:

This will not factor we we will use the Quadratic Formula:

Simplifying...




which is short for:
or

Now for the second equation we got earlier:

Before we get the zero, let's simplify the right side:

Adding and subtracting 11x we get:

Again this will not factor so again we use the formula:

Simplifying...




which is short for:
or

So there are four solutions to your equation:
or
or

You can put this solution on YOUR website!

Any quadratic equation can be solved using the Quadratic Formula. Or, since there is no x-term (just x-squared), we can solve this more simply. Adding to each side we get:

Now we find the square root of each side. Since there both positive and negative square roots we end up with two equations:
or
All that's left is to simplify the square root. Since 80 = 16*5 and since 16 is a perfect square, these square roots will simplify:

Substituting the simplified square roots into our solutions we get:
or

In the future, put parentheses around multiple-term numerators and denominators. Without the parentheses what you posted meant:

But I strongly suspect that the actual fraction is:


Believe it or not, simplifying/reducing fractions is the same as it has always been. Any factors that are common to both the numerator and denominator can be canceled. What has changed since the "good old days" of reducing fractions like 10/20 is that it is harder to find the factors.

So to see if the fraction will simplify/reduce is to find the factors. And when factoring we always start with the greatest common factor, GCF. The GCF in the numerator is 2x. The GCF in the denominator is just 1. Factoring the GCF's out we get:

As we can see, we do have a factor in common, 2x, which we can cancel:

leaving:

which simplifies to:

You can put this solution on YOUR website!

Solving equations like this usually starts with transforming the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(expression)

With the "non-log" term of -1 on the right side it will be harder to transform your equation into the second, "all-log" form. So we will aim for the first form. This will require that we find a way to combine the two logs into one.

If they were like terms then we could just subtract them. But like logarithmic terms have the same bases and same arguments. Our logs have the same bases, 1/3, but the arguments are different. So we cannot subtract them.

Another way to combine logs is to one of the following properties:

• 
• 

These properties require the same bases and coefficients of 1. Our logs meet these requirements. We will use the second property because its logs, like ours, have a "-" between them:

[Note: This property changes the subtraction of two logs not into a division of the logs as you asked but into the log of the division of the arguments. This is a subtle but important difference!]

Our equation is now in the first form. The next step with this form is to rewrite the equation in exponential form. In general, is equivalent to . Using this pattern on our equation we get:

Since -1 as an exponent means reciprocal the left side simplifies to:


Now that the x's are out of logarithms we can solve the equation. First let's get rid of the fraction by multiplying each side by :

which simplifies to:

This is a quadratic equation so we want one side to be zero. Subtracting the entire right side from both sides we get:

Now we factor (or use the Quadratic Formula). This factors easily:

From the Zero Product Property:
2x = 0 or x-2 = 0
Solving these we get:
x = 0 or x = 2

Last of all we check. This is not optional when solving logarithmic equations like or this when you multiply both sides of an equation by an expression that might be zero. We have done both! We must at least check that the solutions create valid arguments and bases to the logarithms. Use the original equation to check:

Checking x = 0:

We can already see that both arguments are going to be zero if x = 0. Zero is not a valid argument for logarithms. (Valid arguments are positive.) So we must reject this solution.
Checking x = 0:

Simplifying...


Using the property to combine the logs:


Check! (If you cannot see that this is a true statement then rewrite this in exponential form.)

So the only solution to the equation is: x = 2

You can put this solution on YOUR website!
I hope your equation is:

and not

or

If I am wrong then re-post your equation using parentheses around the arguments of logarithms and around fractions that are factors (like: ).

Solving equations like:

usually starts with transforming the equation into one of the following general forms:
log(expression) = number
or
log(expression) = log(expression)

Our equation is already in the first form. The next step with this form is to rewrite the equation in exponential form. In general, is equivalent to . Using this pattern on our equation we get:


All we have left to do is simplify the left side. As your note indicates, 1/64 to the 3/2 would be:

What your note doesn't say is you can also use the cube of the square root:

Either can be used. So the question is: Which looks easier? Do you want to cube 1/64 and then find a square root? Or find the square root of 1/64 and then cube that? Since the square root of 1/64 is easy I like square root first and then cube:




When solving equations where the variable is in the argument (or base) of a logarithm, you should check your solutions to make sure the arguments (and bases) remain valid. (Valid arguments to logarithms must be positive. Valid bases for logarithms must be positive but not 1.) Use the original equation to check:

Checking x = 1/512:

Our argument and base are valid so our solution passes the required part of the check.

You can put this solution on YOUR website!

Since 1/4 and 16 are both powers of 2, we can figure this out "by hand".

For the numerator, just ask yourself: "What power of 2 is 1/4?" or "2 to what power equals 1/4?" The answer to these equations will be your numerator. Hint:

For the denominator, just ask yourself: "What power of 2 is 16?" or "2 to what power equals 16?" The answer to these equations will be your denominator. This should not be hard to figure out if you just try different powers of 2.

Your final, fully simplified answer should be:


P.S. This may be just a coincidence by your original expression is what you get if you apply the change of base formula:

to change

into base 2 logarithms.

You can put this solution on YOUR website!
Similar triangles have corresponding sides that are proportional. This means that the ratios of each pair of corresponding sides are all equal to each other.

So the ratio of the two shortest sides:
7.5/10 is equal to the ratio of the two middle length sides and to the ratio of the two longest sides. We can use this to find the other sides of the second triangle:
For the middle sides:
(where "M" is the middle side of the second triangle).
Cross-multiplying we get:

Dividing by 7.5:


For the longest sides:
(where "L" is the longest side of the second triangle)
[Note: We could also have used the ratio of the middle sides (now that we know them both):
(where "L" is the longest side of the second triangle)]
Cross-multiplying:

Dividing by 7.5:


Now that we have all three sides of the second triangle we can find its perimeter:
10 + 12 + 14 = 36

You can put this solution on YOUR website!
The elimination method involves:

1. Setting up the equations so that opposite variable terms are lined up. This can be the hardest part. Sometimes terms have to be rearranged to line up like terms. Sometimes one or both of the equations will need to be multiplied by some number to create opposite terms.
2. Adding the two equations together. The lined up opposite variable terms will cancel each other out resulting in an equation with just one variable.
3. Solve the one-variable equation.
4. Use the solution to the one-variable equation to find the second variable. Substitute for the solved variable in one of the original equation and solve for the second variable.
5. If you want/need to check, then substitute both variables into both of the original equations. If either equation fails to check out (i.e. be a true statement) then a mistake was made somewhere.

Let's see this in action.

-7x-3y = 12
9x+3y = -12
1. Line up the opposites
Your y terms, -3y and 3y, are opposites and they are lined up with each other. So are all set.
2. Add the equations.
2x + 0 = 0
3. Solve the one-variable equation.
2x = 0
x = 0
4. Use the solution above to find the other variable. Substituting for x into the first equation we get:
-7(0)-3y = 12
0-3y = 12
-3y = 12
y = -4
5. Check
-7(0)-3(-4) = 12
9(0)+3(-4) = -12
(I'll leave it up to you to finish the check if you want one.)

P.S. Most of the time, when you add the equations only one variable disappears. (In our problem the y terms disappeared.) Sometimes both variables disappear! In other words you only have numbers (no variables) left). In this case the equation is either a true statement (like 34 = 34 or -0.4 = -0.4 or 0 = 0, etc.) or a false statement (like 4 = 3 or 1/2 = 9 or -3 = 3, etc). If the statement is...

• TRUE. This means that the two equations were actually equations for the same line! This means the lines, in effect, lay on top of each other. Every point that fits one equation will fit the other one, too. This means they intersect at every point! In short, this means that there are an infinite number of solutions to the system (not just the usual single ordered pair).
• FALSE. This means the lines are parallel and do not intersect at all. In short, this means there are no solutions to the system.

You can put this solution on YOUR website!
An equation where the variable is in an exponent.

You can put this solution on YOUR website!
The null/empty set is a subset of every set. The other subsets will be all the sets formed from using 1 or more members of the set:
{3}
{5}
{8}
{11}
{3,5}
{3,8}
{3,11}
{5,8}
{5,11}
{8,11}
{3,5,8}
{3,5,8}
{3,8,11}
{5,8,11}
{3,5,8,11} (A set is a subset of itself)

You can put this solution on YOUR website!
What you posted meant:

This is a difficult problem to solve.

But I suspect that what you meant was:

If I am correct, then please put parentheses around multiple-term exponents like x+6. Clearly posted problems get quicker responses. (Also use parentheses around multiple term numerators, denominators, function arguments (like sin(2x-45)), etc. IOW, use parentheses generously to ensure that there is only one correct way to interpret what you post. Tutors more much more likely to respond to problems that are clear.

To solve

we can use logarithms. Logarithms are the usual way to solve equations where the variable is in exponents. But sometimes there is an easier, more exact way to solve these equations. If you can rewrite the equation so that each side of the equation is a power of the same number, then the solution is much easier than with logarithms. So it is worth a little effort to see if this can be done.

Right now the equation says that powers of different numbers, 1/3 and 81, are equal. Is it possible to...

• rewrite 1/3 as a power of 81?
• rewrite 81 as a power of 1/3?
• rewrite both 1/3 and 81 as a power of some third number?

If any of these are possible then we have an easy solution in front of us.

If you are clever and good with exponents then you will see how all three of the above being possible. If you cannot see this immediately, then the third option above, changing both numbers into powers of some third number, can be the easiest to figure out. So the question is: 1/3 and 81 can both be rewritten as powers of what number? With some real thought about exponents it should not take long to get the answer: 3! and . Replacing 1/3 and 81 with these powers of 3 gives us:

Our equation now has powers of powers. The rule for exponents in this situation is to multiply the exponents. Multiplying the exponents we get:

The equation is now in the desired form: two powers of the same number, 3, which are equal. The only way these powers of three can be equal is if the exponents are equal, too. So:

Now that the variables are out of the exponents we can solve. Adding x to each side:

Dividing by -3:


P.S. When solving equations like this one, it is always worth a little time to see if you can solve it the way we did here because this way, even with the effort it took to figure out that 1/3 and 81 were both powers of 3, is much easier than using logarithms. (Also, if you use your calculator to find logs then you will most likely be working with decimal approximations. These approximations could lead to answers like x = 2.0000001 or x = 1.9999998 instead of the x = 2 we got for this problem.)

However there are equations where it is not possible to rewrite the equation as powers of the same number that are equal. For these equations you must use logarithms.

You can put this solution on YOUR website!
This is an exponential growth/decay problem. A general equation for these is:

where
t = number of units of time
A = the amount after t units of time
= the initial amount (IOW: The amount at t = 0)
r = the factor of change for 1 unit of time. If r > 1 then the equation is for growth and if 0 < r < 1 then the equation is for decay.

In your problem:
t = the number of units of 7 seconds that have passed.
A = 6.25
r = 1/2 (since the amount decreases by 1/2 every 7 seconds)
= 100
So your equation is:


Now we solve for t. We start by isolating the base, 1/2, and its exponent by dividing each side by 100:

which simplifies to:

If the left side is a power of 1/2 then we can solve this by hand. If not, then we will need to use exponents. It will easier to see if the left side is a power of 1/2 if we rewrite it as a fraction:

Clearly 25 will go into both 625 and 10000:

Clearly 25 will go again into both 25 and 400:

If we know (or try out) some powers of 1/2 we should quickly find that . So t = 4. Now remember that t is how many sets of 7 seconds that have passed. So the amount reduces from 100 mg to 6.25 mg not in 4 seconds but in 4*7 or 28 seconds.

You can put this solution on YOUR website!
This is essentially an exponential growth/decay problem. A general equation for these is:

where
t = number of units of time
A = the amount after t units of time
= the initial amount (IOW: The amount at t = 0)
r = the factor of change for 1 unit of time. If r > 1 then the equation is for growth and if 0 < r < 1 then the equation is for decay.

In your problem:
n = 30 days
r = 2 (since the amount doubles each day)
= 0.05
So your equation is:

Finding :

Multiplying:

So after 30 days you will have $53,687,091.20 (Close to $54 million dollars!!)

You can put this solution on YOUR website!
Your statement is of the form: (a and b)
The negation of this would be: (a and b)'
which, according to DeMoivre's (sp?) theorem, simplifies to:
(a' or b')
which, translated back to English, would be:
The sun is not shining or it is not warm.