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log(-6.83) does not exist. It is impossible to raise 10, the base of "log", to any power and get a negative result like -6.83.

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Let's start by listing these numbers in an organized way. We may be able to see a pattern that will help us find the answer quickly. (If we don't find a pattern them we can just list them all and count.)

Numbers do not start with a zero (that counts) so we will start with the two-digit numbers that start with 1:

How many    Two-digit numbers
1           10

All the other two-digit numbers that start with 1 will have a units/ones digit that is not less than the tens digit. Now we'll add the numbers that start with 2's and 3's:

How many    Two-digit numbers
1           10
2           20 21
3           30 31 32

At this point we may see a pattern that helps. We can see that the first row has 1 number, the 2nd row has two numbers, the 3rd row has 3 numbers. So our last row, the 9th, will have nine numbers. You may have learned a formula for the sum of consecutive Integers: n*(n+1)/2. Since we will have 9 numbers to add this would be: 9*(9+1)/2 = 9*10/2 = 45.

If you don't know about this formula then picture this:

1. The numbers are forming a right triangle as we go down the list.
2. Imagine another triangle just like the one our list makes.
3. Imagine this 2nd triangle flipped upside down so that the longest row (the numbers in the nineties) is on the top.
4. Now imagine the two triangles next to each other so that the "10" of the first triangle is in front of the 9 nineties numbers from the flipped triangle; the "20 21" is in front of the 8 eighties numbers fron the flipped triangle; and all the way down the the 9 nineties numbers from the first triangle in front of the "10" of the flipped triangle.
5. The combined shape of the two triangles should be a rectangle with 9 rows and 10 numbers in each row. This means there are 9*10 or ninety numbers in this rectangle.
6. But this rectangle has two copies of every number, one from each triangle. So the count of the numbers in just one triangle is going to be 1/2 of the total for the rectangle: 90/2 = 45.
   Note how we just figured out the n*(n+1)/2 formula by using the two triangles and a rectangle like this!

Or you could just add up how many numbers there are in each row:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

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Here's a procedure for solving equations like these:

1. Isolate a square root.
2. Square both sides of the equation. (Be careful with this step. This is where most mistakes are made. Squaring the isolated square root is as easy as it looks. Mistakes are often made when squaring the other side of the equation.)
3. If there are still any square roots remaining, repeat steps 1,2 and 3. (Note: Squaring both sides of an equation correctly does not always eliminate all the square roots. So there may still be square roots after step 2!)
4. At this point there should be no square roots left. Use appropriate techniques to solve whatever kind of equation you now have.)
5. Check your solution(s). This is not optional when solving these equations. Squaring both sides of an equation, which we have done at least once at step 2, can introduce what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. These extraneous solutions can occur even if no mistakes have been made! So you must check to see if your possible solutions actually work.

Let's see this in action:

1. Isolate a square root.
The square root on the right is already isolated.

2. Square both sides.

The right side squares easily. But we must use FOIL or the pattern to square the left side correctly. I prefer using the pattern:

Simplifying the left side:

The 1's on the left cancel:


3. We still have a square root. So back to step 1.
1. Isolate a square root.
There's only one left. Subtracting 2x from each side:

The square root on the left is sufficiently isolated. But if the -2 in front bothers you then divide both sides by -2 to get rid of it.

2. Square both sides.

The left side is fairly easy to square. We'll use the pattern again on the right side:

Simplifying:


3. The square roots are gone! So on to step 4.

4. Solve the equation.
We now have a quadratic equation. To solve it we want one side to be zero. Subtracting 8x and adding 4 we get:

Now we factor:
0 = (9x-5)(x-1)
From the Zero Product Property:
9x-5 = 0 or x-1 = 0
Solving these we get:
x = 5/9 or x = 1

5. Check the solution(s).
Use the original equation to check:

Checking x = 5/9:

Simplifying...






This is not true. This "solution" does not work. It is extraneous and we must reject it as a solution.

Checking x = 1:

Simplifying...




This solution checks out! So the only solution to your equation is:
x = 1

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A decimal is rational if

• it terminates (stops) (like -0.763); or
• it does not terminate but it has a set of infinitely repeating digits (like 4.8134134134...)

A decimal is irrational if it does not terminate and it does not have repeating digits (like or the number "e" or many square roots)

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The technique we will use to factor can be described as "un-FOIL-ing". FOIL is used to multiply expressions like (a+b)(c+d). Since factoring is, in effect, "un-multiplying" an expression we can use the reverse of FOIL to turn into an expression of the form (a+b)(c+d).

If we have done enough FOIL-ing we know that when (a+b)(c+d) results in a trinomial, like , then:

• The first term of the trinomial, , usually comes from the "F" part of FOIL (which would be a*c). For there is really only one possible "a" and "c": 2x and x.
• The third term of the trinomial, , usually comes from the "L" part of FOIL (which would be c*d). For we could have two different c/d pairs: -3y and y or 3y and -y.
• The middle term of the trinomial, -5xy, usually comes from the sum of the "O" and "I" parts of FOIL (which would be a*d + b*c).

Now we just try the different possible values for a, b, c and d to see if any of them will create the proper middle term. (If none do then the trinomial will not factor.) As it turns out, there is a combination that works:
a = 2x, b = y, c = x and d = -3y
So factors into:
(2x + y)(x + (-3y))
or, more simply:
(2x + y)(x - 3y)

To check this, just use FOIL!

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We will use factoring to solve this. But with only two terms, it is pointless (if it is possible at all) to use factoring by grouping to solve this equation.

First get one side to be zero (by subtracting the from each side):

And then we factor. First, as always, is to factor out the greatest common factor (GCF) if it is not a 1. The GCF here is :


Now we try to factor the factor. Among the factoring patterns you learn is the pattern for difference of cubes: . This pattern may be used here since is an obvious cube and 27 is . Using an "x" for the "a" and a "3" for the "b", then pattern tells us how will factor:

which simplifies to:


With the factoring complete we can now solve. From the Zero Product Property:
or or
The first two equations are simple to solve. (We should get x = 0 and x = 3.) For the third equation we must use the quadratic formula:

which simplifies as follows:



With the negative number in the square root we will get only complex number solutions from this equation. If we are not interested in complex number solutions (or if you don't even know about complex numbers) then we stop here and just use the solutions we found from the other two factors: 0 and 3. If we do want complex solutions then we continue:





which is short for:
or
In standard a + bi form this would be:
or

This makes the solutions, including these complex number solutions:
x = 0 or x = 3 or or

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Method 1: One of the patterns you should have learned by now is:

This pattern can be used to multiply with the "a" being "3u" and the "b" being "1":

(Note the parentheses. Always a good idea when using patterns!)


Method 2:
means (3u-1)(3u-1). You can use FOIL to multiply this out. (After you multiply and then add like terms, you get the same answer as method 1 .)

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If we factor a 4 out of 4q+3 we get

So if (4q+3) is a factor of our polynomial then so will . This fact is useful because we can more obviously use synthetic division with the :

-3/4 |   20   23   -10   p
------       -15    -6  12 
        --------------------
         20    8   -16  12+p

In order for (and therefore (4q+3)) to be a factor, the remainder, 12+p, must be zero. So "p" must be -12.

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The problem is to find the lateral surface area of a cylinder. If you know the formula for this you can use it to find the answer.

If you do not know the formula it is still possible to find the answer. You just have to realize that the label on the can is simply a rectangle that has been wrapped wound the can. To find the area of this rectangle we just have to know the length and width (or base and height) of the rectangle.

The height of the can, 15cm, is one side of the rectangle. The other side of the rectangle is the length of the circular edge at the top (or bottom) of the can. This length is the circumference of the circle. Since the formulas for circumference is:
or
Since we have the diameter, d, we will use the first formula:

or


Now that we have the length and width of the rectangle we can find its area:

which simplifies to:


This is an exact solution for the area of the label. Your problem says to round this to the nearest 100th which I will leave up to you. Just replace the with 3.141592693, multiply by 202.5 and then round.

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6 and -5 are the correct solutions.

When you check -5 as a solution you are supposed to replace the x's in:

with a -5. In the first term, the x is being squared. So when we replace it with -5, the -5 should be squared, too. A proper replacement of the x's with 6's and -5's would look like:

and

Notice the parentheses!! They make little difference with the replacements with 6's. But they are critical when replacing the x's with -5! So it is an extremely good habit to use parentheses when making substitutions.

Your mistake was not using the parentheses and then finding
-5^2
which means "the negative of 5 squared" which is, of course, -25.
What you should have been doing was
(-5)^2
which means "the square of -5" which is, of course, 25. With a 25 instead of -25 you will find that -5 does check out.

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First we should recognize that the graph of a quadratic function, like this one, will have a parabola that opens upward (or downward) as a graph. Such a parabola will have a maximum or minimum value.

Since the coefficient of the squared term is negative (the "a") this parabola will open downward. If you picture such a parabola you will realize that the vertex will be the highest (i.e. maximum) value. So f(x) will have a maximum value at its vertex.

To find this maximum value we just have to find the vertex of the parabola. This can be done by completing the square to put the equation into vertex form. But it is a little easier if you know that the x coordinate of the vertex of any quadratic function will be:

Your "a" is -1 and your "b" is 2. So:

which simplifies to:

Now to find the maximum value we find the y coordinate of the vertex, f(1):

Simplifying...




So the vertex of the parabola is (1, 8) and the maximum value for the function is 8.

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Since the region to be enclosed is rectangular we know its area will be length times width (or base times height). If we call the width "x", then the length will be "2x" (since we are told it is twice the length of the width).
We are also told that the area is to be 162 square feet. So the equation we can use is:

Now we solve for x. First we simplify:

Dividing both sides by 2:

Subtracting 81:

Factoring:
(x+9)(x-9) = 0
Using the Zero Product Property we get:
x+9 = 0 or x-9 = 0
Solving these:
x = -9 or x = 9

Since x represents the width of the region, we will discard the negative solution. (We are not interested in negatives sides of a region.) This makes 9 the only useable value for x (which is the width). And the length will be twice as much: 18.

Finally we answer the question: How much fencing should be purchased? Since the barn will serve as one side of the region we will no need fence for that side. We only need fencing for 1 length and 2 widths:
18 + 2*9 = 18 + 18 = 36 feet of fencing will be required.

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The formula for the area of a parallelogram is: A = b*h. The base in these area formulas can be any side of the polygon. Since all the sides of this polygon are 4's then we know our base will be 4.

The hard part is the height. If you draw a diagram of what you describe then you will see that an equilateral triangle with sides of 4 is formed by two sides of the parallelogram (rhombus actually) and the diagonal whose length is 4. The height of this triangle will also be the height of the parallelogram.

The angles in all equilateral triangles are always 60 degrees. If we draw in the height for the triangle we will have a 30-60-90 right triangle with a hypotenuse of 4. From Trigonometry or from remembering the relationships between the sides of 30-60-90 triangles we should be able to figure out that height is

So the area of this parallelogram will be:

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• First, what are you/we supposed to do with this?
• Second, is the function:
  
  (which is what you posted) or something else like:
  
  If the function is anything other than the first one then put parentheses around the entire numerator and around the entire denominator.

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Let x = the shorter leg. Then the longer leg would be (x + 17) and the hypotenuse would be (3x + 4). In order for this to be a right triangle these sides must fit the Pythagorean equation:

Now we solve for x. First we simplify. (Be sure to use FOIL or the pattern to square the x+17 and the 3x+4.)



Since this is a quadratic equation we want one side to be zero. Subtracting the entire left side from each side we get:

This will factor, but not easily. So you may prefer to use the Quadratic Formula instead.

From the Zero Product Property:
7x + 39 = 0 or x - 7 = 0
Solving these we get:
x = -39/7 or x = 7
We will reject the first solution not because it is a fraction but because it is negative. x is the length of the shorter leg and we cannot have negative lengths.

So the shorter leg is 7 cm.
The longer leg would be x + 17 = 7 + 17 = 24 cm
And the hypotenuse would be 3x + 4 = 3(7) + 4 = 21 + 4 = 25 cm

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Since side CA is adjacent to angle A and side AB is the hypotenuse. The Trig functions that have adjacent and hypotenuse in their ratios are cos and sec. Since our calculators probably do not have a sec button we will use cos:
where x is the length of side AB
Multiplying each side be x we can eliminate the fraction:

Dividing each side by cos(55):

This is an exact expression for the solution. For a decimal approximation we use our calculators to divide 10 by cos(55):


Rounded to the nearest integer this would be 17.

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First let's look at the graph of the two parabolas:

The red one is and the green one is . The solution to would be the red parabola and all the area above/inside the bowl. The solution to would be all the area below/inside the green parabola. (Not the green parabola itself because that is where y equals and the inequality does not include "or equal to".)

So the solution to the system is where these two areas overlap each other: The enclosed area between the two parabolas. To express this solution we must first find the two points where the parabolas intersect. Setting

and solving for x we should be able to find the points of intersection. Subtracting the entire right side we get:

Factor out the GCF of 2:

Factoring the trinomial:

From the Zero Product Property:
x + 1 = 0 or x + 2 = 0
Solving:
x = -1 or x = -2

Using these x values and either equation for the parabola we can find that the y values for each x are 1 and 2, respectively. So the points of intersection are:
(-1, 1) and (-2, 2)
So the x values of the solution are between -1 and -2, inclusive:

and the y values of the solution are:

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The problem tells us nothing about any zeros or factors so we will have to find them on our own. Always start factoring by factoring our the greatest common factor (unless it is a 1 which we rarely bother factoring out). The GCF is a 1 so we will not factor it out.

Next we usually try factoring by patterns or trinomial factoring because they are often easier than the other two factoring techniques. However out expression has too many terms for a trinomial or for any of the patterns.

Now we are left with factoring by grouping or with factoring by trial and error of the possible rational roots. Factoring by grouping requires an even number of terms. Since we have 5 terms we would have to split the into two parts before we try to use factoring by grouping. Since I do not see which particular split (, , , etc.) I am going to try to avoid using this method an go on to factoring by trial and error of the possible rational roots.

The possible rational roots of a polynomial are all the ratios, positive and negative, which can be formed by a factor of the constant term (at the end) over a factor of the leading coefficient (at the beginning). Our constant term is 64 and our leading coefficient is 3. The factors of 64 are 1, 2, 4, 8, 16, 32, 64 and the factors of 3 are just 1 and 3. So the possible rational roots of f are:
+1/1 (or +1), +2/1 (or +2), +4/1 (or +4), +8/1 (or +8), +16/1 (or +16), +32/1 (or +32), +64/1 (or +64), +1/3, +2/3, +4/3, +8/3, +16/3, +32/3, +64/3

With so many possible rational roots it could take quite a while to find some roots. It will help if we use our brains to eliminate some of the possible roots logically. With only two terms with negative coefficients, it is unlikely the larger positive roots could make f be zero. So we will not try 4, 8, 16, 32 or 64 (unless nothing else works). Let's try 2 (using synthetic division which I hope you have learned):

2  |    3   -19   54   176   -64
----          6  -26    56   264
       --------------------------
        3   -13   28   232   200

The 200 in the lower right corner is the remainder. The remainder is f(2). Since f(2) is not zero, 2 is not a root. (And the fact that it is a large positive number tells us that even 2 was too large of a positive root to work.)

We can try 1 and -1 mentally since powers of 1 and -1 are easy:
f(1) = 3 - 19 + 54 + 176 -64 = 150.
f(-1) = 3 + 19 + 54 - 176 -64 = -164
Neither of them work. But the fact that f(1) is positive and f(-1) is negative tells us that there is a root between them. Since zero is nearly halfway between 150 and -164 and since 150 is a little closer to zero, I'm going to guess that 1/3 might be a root. Let's see:

1/3 |    3   -19   54   176   -64
----           1   -6    16    64
        --------------------------
         3   -18   48   192     0

Bingo! The remainder is zero so f(1/3) = 0 and (x - 1/3) is a factor. Not only that, the rest of the bottom line is the other factor of f. "3 -18 48 192" translates into: . So:

We can simplify this a little if we factor out a 3 from the second factor:

and then multiply (x - 1/3) by the 3:


Now we continue to factor. The last factor, , has the same constant term, 64, but the leading coefficient is now a 1. So we have the same list of possible rational roots as before except for the fractions. And since a rational root that didn't work before (-1, 1, 2, 4, 8, 16. 32 and 64) cannot magically start working later we only have -2, -4, -8, -16, -32 and -64 left to try. Let's start with -2:

-2 |  1   -6   16   64
----      -2   16  -64   
     ------------------
      1   -8   32    0

The remainder is zero. So f(-2) = 0 and (x - (-2)) or (x + 2) is a factor. And the other factor is :


The last factor is a quadratic. So we can use alternate factoring techniques. But no matter what technique we use, it will not factor. But we can use the quadratic formula to find its roots:

which simplifies as follows:










which is short for:
or
With these two roots we can write the remaining factors:
(x - (4 + 4i)) and (x - (4 - 4i))
which simplify to:
(x - 4 - 4i)) and (x - 4 + 4i))
Adding these two factors to our factored f:

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First of all, please try to post your questions in an appropriate category. You will get a quicker response if you do so. This problem has nothing to do with quadratic equations. I have changed it to an appropriate category.


Your work is all good so far. And you may actually be finished. It depends on the exact meaning of "expand". For many teachers they want you to make the arguments as simple as possible. If this is the case then you need to use one more rule, , to move the exponent of the "c" out in front:

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100% correct!

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I assume the equation is:

If I am correct, then please put parentheses around exponents that are not just a number or variable. What you posted meant:
(Note the first x is not in the exponent.)

Since the first exponent on 5, 2x, is exactly twice the exponent on the second 5, x, this equation is in what is called quadratic form. These equations can be solved using the same techniques used on regular quadratic equations.

If you have not solved many of these they can be a bit difficult. It can be helpful to use a temporary variable:
Let . Then , Substituting these into our equation we get:

This is obviously a quadratic equation. It factors fairly easily:

From the Zero Product Property:
q - 7 = 0 or q - 2 = 0
Solving these we get:
q = 7 or q = 2

Of course we are interested in values for x, not q. So we substitute back in for q:
or
To solve these for x we will use logarithms. Any base of logarithm may be used. But there are some advantages to certain bases:

• Choosing a base of logarithm that matches the base of the exponent will result in the simplest possible expression for the solution.
• Choosing a base of logarithm that your calculator "knows" (base 10, "log", or base e, "ln") will result in an expression that can be easily converted to a decimal approximation if one is needed/wanted.

I'm going to match the bases so I'll use base 5 logs:
or

Next we use a property of logarithms, , which allows us to move the exponent of the argument out in front:
or
By definition . So these equations simplify to:
or
These are exact expressions for the solutions to your equation.

If you want/need a decimal approximation, then use the change of base formula, , to convert this to base 10 or base e logs so you can use your calculator.

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I'm guessing that the reason no one has responded by now is that we cannot tell if the equation is:
sin(2x)+cos(2x)+tan(2x)+cot(2x)+sec(2x)+csc(2x)=7
or

or some variation of these.

Please re-post your question and

• Put parentheses around arguments to functions; and
• Use "^" to indicate any exponents.

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The expression in the parentheses is one of the variations of cos(2x):

The second term is the square of sin(2x):


which equals
1

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You had a good idea with replacing the denominators. Replacing sec(x) with 1/cos(x) and tan(x) with 1/cot(x) we end up with:

Since the right side is a product we want to be able to write the right side as a product, too. So we are looking to factor the left side. While the left side is a difference of squares, factoring it that way does not (as far as I can tell) get us closer to the end of the proof. Instead we should know that . Looking at this way we can see that
If we're really cleaver we would factor out since that is a factor of the right side:

We can replace the fraction with :

From one of the Pythagorean identities :

which simplifies to:

And we are finished.

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

If we can find a way to combine the two logarithms into one, then we would have the first form. There are two ways to combine log terms:

• Adding (or subtracting) them. This requires that the logarithms have the same bases and the same arguments.
• Use a property of logarithms:
  • when the terms has a "+" between them
  • when the terms has a "-" between them
  Both of these properties require that the logarithms have the same base and coefficients of 1.

Since both of these methods require logarithms of the same base and since our logs have different bases, we start by changing the base of one or both logs so that they are the same. To change bases we use the Change of Base formula: . We need to change the base 9 log into an expression of base 3 logs or vice versa or change both bases into the same third number. Since 9 is well-known power of 2 (and vice versa for that matter) we will not need to change both bases. We can just change base 9 into base 3. (I'll show changing base 3 into base 9 at the end.) Using the change of base formula to change the base 9 log:

Since :

To get rid of the fraction I'll multiply each side by 2:


Now that the bases are equal we can start to combine the terms. The arguments are different so they are not like terms. So we cannot subtract them from each other.

The properties require coefficients of 1 and the second log has a coefficient of 2. So we cannot use the properties, yet. Fortunately there is another property of logarithms, , which allows use to move a coefficient into the argument as its exponent. Using this property we can move the coefficient of 2:

which simplifies to:

The logs now meet the requirements for the properties that combine terms. We'll use the second one because its logs, like ours, have a "-" between them:

We finally have the first form.

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

which simplifies to:


Now that the variable is out of the logs we can solve. Multiplying by 4x:

Subtracting 4x:

Adding 3:
Last of all we check. This is not optional when solving logarithmic equations. You must at least ensure that all argument remain valid (i.e. positive) for each solution. Any "solution" that makes an argument invalid (zero or negative) must be rejected.

Use the original equation to check:

Checking x = 3:

We can already see that both arguments will be positive when x = 3. So this solution passes the check. (If x = 3 failed this check we would reject it and, since it was the only "solution" we found, it would mean that there were no solutions to the equation.)

P.S. Changing the base 3 log into base 9:


Since 3 is the square root of 9 and since an exponent of 1/2 means square root, the denominator is equal to 1/2:

which simplifies to:

This rest is very similar to what we did above and we end up witht he same solution.

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What you posted means

But for various reasons I suspect that what you meant was:

If I am correct, then please use parentheses to clarify your expressions. For example:
2/5=4/(1+e^(6x))
If I am wrong about your equation then the rest of this will not help you.

First let's eliminate the fractions by multiplying both sides by the lowest common denominator, :

which simplifies to:


Next let's isolate the base and its exponent. Subtracting 2:

Divide by 2:


Now we use logarithms. Any base of logarithms may be used. But there are advantages to choosing certain bases:

• Choosing a base of logarithm that matches the base of the exponent will result in the simplest possible expression for the solution.
• Choosing a base of logarithm that your calculator "knows" (base 10, "log", or base e, "ln") will result in an expression that will be easy to convert into a decimal approximation.

In this equation, with the base of the exponent being e, we can get both advantages by using ln:

Next we use a property of logarithms, , which let's us move the exponent of the argument out in front. (It is this very property that is the reason we use logarithms on equations like this. It give us a way to move the exponent, where the variable is, out in front where we can then "get at" the variable with "regular" algebra.) Using this property we get:

By definition ln(e) = 1. (This is why matching the bases of the exponent and logarithm leads to the simplest possible expression for the solution.)

Now we can solve for x. Dividing by 6:

This is an exact expression for the solution to your equation. You were also asked to find a rounded decimal approximation for the answer. I'll leave that up to you and your calculator. (Just make sure you find ln(9) first and then divide by 6.)

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

If we could find a way to combine the two logarithms we would have the first form. They are not like terms so we cannot subtract them. (Like logarithmic terms have the same bases and the same arguments.)

But there is a property of logarithms, , which provides an alternate way to combine logs which have a "-" between them. This property requires the same bases and coefficients of 1. Our logs meet both requirements. Using this property on the left side of our equation we get:

We now have the desired form.

The next step with this form is to rewrite it in exponential form. In general is equivalent to . Using this pattern (and the fact that the base of "log" is 10) we get:

which simplifies to:

Now we solve for x. Multiplying both sides by x we get:

Adding x:

Dividing by 11:


Last we check. This is not optional when solving equations like this. You must at least ensure that the arguments of the logs are valid (i.e. positive). Use the original equation to check:

Checking x = 3/11:

We can already see that the first argument, 3 - 3/11, will turn out negative. An argument to a logarithm may never be negative. So we must reject this solution. And since this rejected "solution" was the only one we found...
There are no solutions to your equation!

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First we isolate the base and its exponent. Subtracting 6 we get:

Dividing by -2:


Now we use logarithms. Logarithms of any base may be used. But there are advantages to choosing certain bases:

• Choosing a base of logarithm that matches the base of the exponent will result in the simplest possible expression for the solution.
• Choosing a base of logarithm that your calculator "knows" (base 10, "log", or base e, "ln") will result in an expression which will be easy to convert to a decimal approximation.

In this equation the base of the exponent is a base our calculator "knows" so we can get both benefits by choosing base e logarithms:

On the left side we can use a property of logarithms, , which allows us to move the exponent of the argument out in front. (It is this very property that is the reason we use logarithms on equations like this. It lets use move the exponent, where the variable is, out in front where we can "get at" the variable with "regular" algebra.) Using this property we get:

By definition ln(e) = 1. (This is why matching the base of the logarithm to the base of the exponent results in the simplest expression.)

Now we can solve for x. Adding 5:

Dividing by 9.4:

This is an exact expression for the solution to your equation. If you want/need a decimal approximation get out your calculator and use it on the right side.

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If the equation is
3*tan(3θ) - 3tan(2θ) - tan(θ) + 1 = 0
then please re-post it with the parentheses.

If the equation is:

then please re-post it with the parentheses and using "^" to indicate exponents:
3*tan^3(θ) - 3tan^2(θ) - tan(θ) + 1 = 0

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A diagram may help make sense of this:

Angle BCD is 90 degrees. This makes triangles BAC and BDC right triangles.
Angle BAD is 23 degrees 10 minutes (or 23.1666 degrees).
Angle BDC is 32 degrees 17 minutes (or 32.2833 degrees).
AD is 1 mile.
BC is the height of the mountain. Let's call it "x".
We will need to refer to DC also. Let's call it "y".

In triangle BAC the vertical side, x, is opposite to angle BAC and the horizontal side, 1+y, is adjacent. Since tan is the ratio of opposite over adjacent:


In triangle BDC the vertical side, x, is opposite to angle BAC and the horizontal side, y, is adjacent. Since tan is the ratio of opposite over adjacent:


Solving the last equation for y...



Substituting this expression for y into the first equation:

Now we solve for x. Multiplying the numerator and denominator of the fraction on the right side by tan(32.2833):

which simplifies to:

Multiplying both sides by the denominator on the right side:

which simplifies to:

Next we gather the terms with x on the right side by subtracting x*tan(23.1666) from both sides:

Factoring out x on the right side:

Then we divide both sides by (tan(32.2833) - tan(23.1666)):

Last of all we use our calculators to find the two tan's:



This is the height of the mountain, in miles. To get the height in feet, just multiply this by 5280.

Note: The angles, the tan's and some of the calculations were all rounded to four decimal places. So there is a small error in the answer. If you round to more than just 4 decimal places then you will get a more accurate answer.

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is a special angle. We should know that its sin is
is also special angle. (Any integer multiple of is a special angle.) The reference angle is and the cos of this is . Since is in the second quadrant and cos is negative there,
is also special angle. (Any integer multiple of is a special angle.) The reference angle is and the tan of this is 1. Since is in the second quadrant and tan is negative there,
is a special angle. We should know that its cot is 1.
Substituting in all these values into your expression we get:

which simplifies as follows: