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When the variable is in an exponent like in this equation, logarithms are often used to solve the equation. But if the equation can be rewritten so that each side is a power of the same number there is a much shorter, easier way. It is worth the effort to see if this is possible.

So we will start by seeing if we can rewrite the equation so that each side is a power of the same number. The left side is already a power of 3. Can the right side be rewritten as a power of 3? It should take long to find that . And since , . So we can rewrite the equation as:

We now have the desired form.

With this form we have an equation that tells us that 3 to the x power is equal to 3 to the -3 power. The only way powers of 3 can be equal is if their exponents are equal. So:
x = -3
And we're done!!

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Substituting in the values given for b and h we get:

And now we solve for p. First we'll isolate the log. Multiplying both sides by -9/100:


Next we rewrite the equation in exponential form. In general is equivalent to . Using this pattern (and the fact that the base of "log" is 10) on our equation we get:

Next we multiply both sides by 31:

This is an exact expression for the solution. I'll leave it up to you to use your calculator to find the decimal. (Just be sure to raise 10 to the -.243 power before you multiply by 31.)

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While long division is always an option for division of polynomials, synthetic division can be used if the divisor is of the form: x - n. Since synthetic division is usually easier that is how I will do this. I hope you've been taught this method of division. (If not, ask your teacher to teach it!)

 2 |   7   0   -8   0   7   0   5   <== Note the 0's for the missing terms
----      14   28  40 160 334 668 
      ----------------------------
       7  14   20  80 167 334 673

The last number in the bottom row is the remainder. The rest of the bottom row is the quotient. The numbers "7 14 20 80 167 334" translate into:
. And the remainder, as usual, goes into the numerator of a fraction with the divisor as the denominator. So:

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Your error was in squaring . According to one rule of exponents, you raise a single term, like , to a power by raising each factor to that power. So according to this rule:

becomes

Squaring the -5 is easy. To square the powers of m and n we use another rule of exponents: Multiply the exponents! (This is where you went wrong. You apparently squared the 3 and got 9 and squared the 2 and got 4.) So we should get:

With properly simplified to we can now multiply it by giving us:


P.S. Your answer has n to the right power. But it is only a lucky accident that squaring the exponent of 2 (which you should not have done) works out the same as multiplying the exponents of 2 (which you should have done)!

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Pascal's triangle provides coefficients for the terms in the expansion of powers of a binomial (like ). Before we figure out the coefficients we are going to look at the terms.

The pattern of the terms, without the coefficients, for ) will be:

Note how the exponents of each term add up to 5. Note also the alternating operations (positive/add then subtract then add ...). The alternating operations are due to the "-" between the 2x and the 5y in . (If there had been a "+" between the 2x and 5y then there would only be additions.)

We can simplify our coefficient-less expression by raising everything to their powers ...

and then multiplying within the terms...


Now we're ready for the coefficients from Pascal's Triangle. Pascal's Triangle starts with:

                                   1
                                 1   1
                               1   2   1
                             1   3   3   1

As you can see, the outermost edges of the triangle are all 1's. For the numbers on the inside you add the number above and to the left and the number above and to the right. For example, in the 4th row:

• The row starts with a 1 because all rows start with a 1.
• The first 3 in the 4th row comes from adding the 1 that starts the 3rd row and the 2 in the 3rd row.
• The second 3 in the 4th row comes from adding the 2 in the 3rd row and the 1 at the end of the 3rd row.
• The row ends with a 1 because all rows end in a 1.

In the 5th row...

• It start's with a 1
• The next number will be the sum of the first 1 in the 4th row and the first 3 in the 4th row.
• Next in the 5th row will be the sum of the two 3's in the 4th row.
• Next will be the sum of the second 3 and the last 1
• And the row ends with a 1.

I'll leave it up to you to figure out the 5th and 6th rows.

Once you have the 6th row you will have a row with six numbers. Earlier we came up with the expression:


which has six terms! Take each of the numbers from the 6th row, in order, and put one of them in front of each of the 6 terms of the expression above. For example the 1 at the start of the 6th row would go in front of the first term:

The second number in the 6th row works out to be 5. So we put this 5 in front of the second term:
(There will still be a "-" in front of this term.)
The 1 at the end of the 6th row goes in front of the last term:
(There will still be a "-" in front of this term.)
Insert the rest of the 6th row in front of the rest of the terms. And last of all, multiply these numbers from the 6th row with each of the terms. So the first, second and last terms work out to be:

(There will still be a "-" in front of this term.)
(There will still be a "-" in front of this term.)

So you end up with:
+ ? - ? + ?
The three ?'s will be replaced by the terms you figured out. (Make sure you keep the alternating signs!)

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:
This is not so much a problem of division but a problem of rationalizing a denominator. The two square roots in the denominator are irrational numbers. Proper form for fractions is to have a rational denominator. So our problem is to find a way to turn the denominator into a rational number.

It is a little harder to do this with two terms in the denominator (compared to one term). The "trick" is to take advantage of the pattern:

One way to look at this pattern is that it shows us how to turn a binomial (an a+b or a-b) and turn it into an expression of perfect squares. This shoes us how to take a binomial with square roots and turn it into an expression of the squares of those square roots. Since our denominator has a "-" between the terms, it will play the role of a-b with the "a" being and the "b" being . To turn it into an expression of perfect squares we need to multiply it by a+b. And of course if we multiply the denominator by something we need to multiply the numerator by the same thing. So we'll multiply the numerator and denominator by :

On top we will simply use the Distributive Property to multiply. On the bottom, the pattern tells us how it will work out. (FOIL can be used but it is slower.)

The squared square roots will simplify:

We now have the rational denominator we wanted. Continuing to simplify...


This is the simplified expression.

P.S. In this problem, the fraction disappeared after we rationalized the denominator. This will not always happen. Often the fraction you have at the end will have square roots in the numerator. For reasons I cannot explain, irrational numbers like square roots are OK in numerators but not OK in denominators.

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Since I cannot see the graph myself I can only tell you what to look for. On the graph of a polynomial, the roots will be the x-coordinates of points where the graph intersects the x-axis, IOW x-intercepts. These will be the real roots of the polynomial. Complex roots cannot be determined from the graph. If the graph never intersects the x-axis then it has no real roots. (A cubic polynomial, like yours, should intersect the graph 1 to 3 three times.)

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1) To find the least common multiple (LCM) you must know what the factors of the expressions are. So we start by factoring (6+5v), (36-25v^2)and (6-5v)

(6+5v)       = 1 * (6+5v)
(36-25v^2)   = 1 * (6+5v) * (6-5v)  [A difference of squares]
(6-5v)       = 1 *          (6-5v)
LCM (or LCD) = 1 * (6+5v) * (6-5v)

I've factored these expressions in this way on purpose. Using spacing, I ensured that the factors in each column were the same. (I'll explain the 1's later.) With the factors arranged this way, the LCM will be the product of the factors from each column. Use only one factor from each column. The LCM here would be:
1 * (6+5v) * (6-5v) = 36-25v^2

2) Expressions can be undefined for a variety of reasons. But for the expression 7/2V+5 there is only one reason: a zero denominator. 7/2V+5 will be undefined for any value of V that makes the denominator zero. Since I cannot tell if the denominator is 2V or 2V+5 you will have to figure this out yourself. Just set the denominator, whatever it is, equal to zero and solve for V. That will tell you the value of V that will make the expression undefined. [If the denominator is 2V+5 as I suspect, then please use parentheses around multiple-term denominators (or numerators): 7/(2V+5)]

P.S. The way I arranged the factors to find the LCM can also be used to find greatest common factors (GCF's). The 1's in the list of factors are not needed to find LCM's but they can be necessary to find a GCF. To use the table of factors above to find the GCF of those three expressions, you use factors that are present in every row. In this case, only the 1's are present in every row. So the GCF is a 1. (If there had been other factors in every row then the GCF would be the product of the 1 and these other factors.)

P.P.S. Least common denominators (LCD's) are simply the LCM of some denominators so you can use the table of factors to find LCD's, too. A frequent use of LCD's is to be able to add or subtract fractions. In this case using a table of factors like the one above can not only tell you what the LCD is but it can also be used to tell you what each fraction needs to be multiplied by to make its denominator equal to the LCD. For example, let's say you wanted to add:

Using the table above we would find that the LCD is 36-25v^2. And it is perhaps obvious that the second fraction is fine as it is. But for the other fractions we have to change them so that their denominators match the LCD. Looking at the table we can compare the factors of each denominator with the factors of the LCM/LCD and see what is missing. The first fraction's denominator is "missing" a factor of (6-5v) and the last fraction's denominator is "missing" a factor of (6+5v). This tells us what to multiply each fraction by:

etc.

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In general, a polynomial that has r as a root will have a factor of (x-r). So the polynomial we are looking for will have factors of:

which simplifies to:


Now all we have to is multiply this out. We can take advantage of the pattern to multiply the first two factors. Or we can take advantage of the pattern to multiply the last two factors. (Remember, with multiplication the order does not matter!) Since the second pattern has a simpler result I'm going to use it to multiply the last two factors:

which simplifies to:

Now we multiply the remaining factors. They do not fit any pattern so we must resort to FOIL:

which simplifies...

Rearranging the terms so they are in standard form:


P.S. When I described the factored form of a polynomial, I left out a minor detail. The general factored form of a polynomial is:
P(x) = a * (x-r) ... (with as many (x-r) factors as there are roots).
The "a" can be any non-zero number and it is the part I left out. The answer we got while ignoring the "a" has, in effect, an "a" of 1. Fell free to re-do the solution and pick a different "a".

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Simplifying square roots involves trying to find perfect square factors of the radicand. (The expression inside a radical is called a radicand.) So we start by factoring the radicand.

The radicand fits the pattern of one of our factoring patterns:

The first term of the pattern is a perfect square. The first term of our radicand is also a perfect square: . The last term of the pattern is also a perfect square. The last term of our radicand is a perfect square, 3 squared. The middle term of the pattern, 2ab, is 2 times what is being squared in the first term time what is being squared in the last term. The middle term of our radicand is equal to 2 times 5x (what is being squared in front) times 3 (what is being squared in back. So the radicand does fit the pattern with "a" being "5x" and "b" being 3. So according to the pattern we the radicand will factor into :


As we can see, not only do we have a factor that is a perfect square, the whole radicand turned out to be a perfect square!

An easy temptation is to think that the square root simplifies to just 5x+3. But every positive number has two square roots, a positive one and a negative one. A square root without a minus in front of it is a reference to the positive square root. So our original square root, , is a reference to the positive square root of . So when we are done simplifying we must have an expression that is just as guaranteed to be positive as the original square root! We are told that x could represent any value. So 5x+3 could be any value, too, including negative. To ensure the "positiveness" of our simplified answer we use absolute value. So

simplifies to:

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If the expression is:

then please post it with parentheses around multiple-term numerators and denominators, like:
(w^2 +2w-24)/(w^2+w-30) +8/(w-5)
What you posted, without the parentheses, meant:

Clearly posted questions will get faster responses.


To add these (or any) fractions we must first make the denominators be the same. To make the denominators the same we first look for the lowest common denominator (LCD). To find the LCD we must see what the factors of the denominators are:


This makes the LCD:
(w+6)(w-5)

We could go ahead and make the second fraction's denominator equal to the LCD. But if we are a little clever and look at the first term's numerator we will notice that it factors, too:

Since the first fraction's numerator and denominator share a factor, w+6. So it will reduce. Reducing the first fraction will make things easier so that is where we will start:


Canceling the common factor:

leaving:


Not only did the first fraction reduce, but the denominators ended up being equal! So we can go directly to the addition:

which simplifies to:

This is the simplified sum.

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Please put multiple-term numerators and denominators in parentheses. What you posted meant:

But I suspect that this is not the actual equation. Please re-post using parentheses to make the numerators and denominators clear. (If you're not sure when to use parentheses, use them all the time! Too many do not cause a problem (as long as they are balanced) while too few often says the wrong thing.)

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There are a few problems with what you posted:

• Every equation you posted is ambiguous. None of them make clear what is inside the square root. For example, the first equation could be any of the following:
  
  
  
  
  You have parentheses around the words "square root". Putting parentheses around the radicand (the expression within a radical is called a radicand) is much more important.
• I don't think "extraneous equations" is a meaningful phrase. Solving square root equations like these can result in what are called "extraneous solutions".

Please re-post your question using parentheses to make the equations unambiguous.

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This is easy if you understand what logarithms are. In general, logarithms are exponents. Base 3 logarithms, like the one in this problem, are exponents one would put on a 3. This specific logarithm, , is the exponent one would put on a 3 to get a result of 9. So what power of 3 results in 9? The answer to that question is what is equal to.

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I just solved this very problem. Click here to see the solution.

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If our 4 roots are , , , then our polynomial, in factored form, will be:

where "a" can be any non-zero number. To find the polynomial we will:

1. Pick the 4 roots
2. Pick a number for "a"
3. Multiply out the factored form and then simplify.

1. Pick the roots.
This will be the hardest part. Lets say that and will be our irrational roots of the polynomial. Most of the irrational numbers you know are roots (square, cube, 4th, etc.). To get integer coefficients the problem specifies, we need the two factors with these roots, , to multiply in such a way that the radicals disappear. is the product of two binomials (two-term expression). So we want the product of two binomials which include radicals to result in something where there are no radicals. The key to figuring this out is the factoring pattern:

This shows us how two binomials can be multiplied and result in an expression of perfect squares. If we make our irrational roots be square roots then we can get rid of the radicals by using this pattern to get perfect square terms. But which square roots? Since the factors of our polynomial are written as subtractions, let's rewrite the pattern as subtractions:

So we want our square roots to be "-b" and "b". In other words, we want square roots that are opposites. So we can use and , or and , or and , etc. Let's pick and . So our polynomial, at this point is:

or


The imaginary roots of the polynomial will have an "i" in them. "i", as you should know, is . Since i is a square root, we can use the exact same logic for picking these roots as we did for the irrational roots. So for the imaginary roots we want opposites, too. Let's pick i and -i. Now our polynomial is:

or


2. Pick a value for "a".
The only rule is that you can't pick zero. Let's be easy on ourselves and pick 1. So our the complete polynomial, in factored form, is:


or


3. Multiply.
Multiplying the first two factors is easy since we can use the pattern:

which simplifies to:

We can multiply the last two factors using the pattern, too. (Remember, with multiplication you can multiply in any order you choose.)

which simplifies to:

Since this further simplifies:


One more multiplication. No pattern can be used. We must use FOIL:

which simplifies:


or

This is a 4th degree polynomial with integer coefficients with two irrational roots, and , and two imaginary roots, i and -i.

You can check each root by replacing the x with that root and see if P(x) is zero. If it is then that root is good.

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What question?? What are you/we supposed to do with that expression?

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Starting with the graph of the "base" function of log(x):

1. The "-" in front of the log will cause a reflection of the graph in the x-axis.
2. The "+2" will cause a vertical translation, up 2.

The vertical asymptote, reflected in the x-axis (a vertical reflection) and then translated up 2, will still be the same! So the vertical asymptote of log(x), x = 0, will also be the vertical asymptote of g(x).

An x-intercept by definition is on the x-axis. All points on the x-axis have a y coordinate of 0. So to find an x-intercept make the y be 0 and solve for x:

Subtract 2:

Divide (or multiply) by -1:

Since the base of "log" is 10 this equation tells us that x is what you get if you raise 10 to the 2nd power, i.e. 100:

So the x-intercept is (100, 0).

Here's a look at the graphs of log(x) (in red), -log(x) (in green) and -log(x)+2 (in blue). Note the transformations. Also, Algebra.com's graphing software is not perfect. All three graphs look like they intersect the y-axis. They do not. The y-axis, x=0, is the vertical asymptote!

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Help you do what? What are you/we supposed to do with f(x)?

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The first key to solving this equation is to recognize that one base with an exponent x, 4, is a power of the other base with an exponent, 2. So we can rewrite 4 as a power of 2:

The first term is a power of a power. The rule for the exponents is to multiply them:


The next key is to notice that the exponent of the first 2 is exactly twice the exponent of the other 2. An equation like this is called an equation in "quadratic form". Quadratic form equations can be solved in much the same way as regular quadratic equations.

If you have trouble seeing the "quadratic-ness" of the equation it can be helpful to use a temporary variable. Set the temporary variable equal to the base and the smaller exponent. So:
Let
Then
Substituting these into the equation we get:

This is obviously a quadratic equation. We can solve it by factoring:

From the Zero Product Property:
or
Solving these we get:
or

Of course we are not interested in solutions for q. We want solutions for x. So we substitute back in for q:
or
To solve for x we have some more work to do. I hope the solution to the first equation is obvious, x=2. If not, then you can solve it in a way similar to the way we will solve the second equation.

Since we do not know what power of 2 results in a 3, we will need to use logarithms to solve for x. The logarithm we use can be of any base. But choosing certain bases have advantages:

• Choosing the logarithm's base to match the base of the exponent (where the variable is) will result in the simplest possible expression for the exact solution.
• Choosing the logarithm's base to match a base our calculators "know" (base 10, "log", or base e, "ln") will result in an exact expression which can easily converted to a decimal approximation is wanted/needed.

I'll do it both ways so you can see the difference.

Matching the logarithm's base to the exponent's base, we'll use base 2 logs:

Next we use a property of logarithms, , which allows us to move the exponent of the argument of a log out in front as a coefficient. (It is this very property of logarithms which is the reason we use logarithms. It allows us to, in effect, change an exponent (where the variable is) into a coefficient (where we can then "get at" the variable with "regular" algebra.) Using this property we get:

By definition, . (This is why matching the base of the log to the base of the exponent results in the simplest expression.) So the left side becomes:

This is an exact expression for the second solution to your equation. (The first solution was x = 2.)

P.S. Using a base that our calculators "know". I'll use base e, "ln". The steps are mostly the same so I'll omit commentary except to explain differences:



This time the log on the left is not a 1. Dividing by the log on the left:

Although it looks quite different from the solution we got using base 2 logs, , this is also an exact expression for the second solution to your equation. By using ln's we get an expression which is not as simple as the other but which can be easily entered into our calculators to get a decimal approximation if wanted/needed.

P.P.S. Once you've gotten some experience with these quadratic form equations you will no longer need to use a temporary variable. You will be able to see how to do directly from:

to

to
or
etc.

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Some logs can be evaluated "by hand" (without a calculator). For these logs it is better to evaluate them by hand because:

• The answer will be exact when you do it by hand. When you use a calculator you often get either a decimal approximation or the decimal version of some simple fraction.
• Doing logs by hand exercises and strengthens your understanding of what logs are.
• Your calculators usually know how to do 1 or 2 types of logarithms:
  • Base 10 logs, "log"
  • Base e (or natural) logarithms, "ln".
  To use a calculator to find logs of any other base, like parts "b" and "c" of your problem, you will have to convert the logs into base 10 or base e logs using the base conversion formula:
  

So we are going to try to do each log by hand first before we reach for our calculator. If we are unable to find the log by hand, we will resort to a calculator.

a) log(19.1)
To do logs by hand you must understand what they are. In general, logarithms are exponents. This particular log is a base 10 log. Base 10 logs, in general, are exponents that one would put on a 10. This specific base 10 log, log(19.1), represents the exponent one would put on a 10 to get a result of 19.1. So to be able to do this log by hand we would have to be able to figure out what power of 10 is equal to 19.1. Since every power of 10 that I know of only has digits of 1's and 0's, I have no idea what power of 10 would result in 19.1. So this is a case for our calculators. Since most calculators have a button for base 10 logs, "log", this should be easy. I get:
log(19.1) = 1.2810333672477275376350435982706
Rounded to two decimal places this would be:
log(19.1) = 1.28

b)
(Note: This is called "the base 2 log of 8" (which is a much better way to describe it than the way you posted this.) This base 2 log is an exponent one would put on a 2 to get a result of 8. If you don't know what power of 2 result in 8, don't immediately reach for your calculator. Explore some powers of 2 first. It should not take long to find that . So:


c)
This base 5 log represents an exponent one would put on a 5. Specifically, this base 5 log is the exponent one would put on a 5 to get a result of 1. This is another one we can do by hand. We should know that any number (except 0) raised to the zero power is equal to 1. So the log of 1, no matter what the base, is always zero:


d) 0.042
???. If this is correct then all we have to do is round to two decimal places:
0.04

P.S. If you are using the calculator program on Windows but you do not see a button for "log" or "ln":

1. Start the calculator program. It is often found under "accessories".
2. Click on the "View" menu.
3. Click on "Scientific" in the list that appears.

This should change the display so that it has many more buttons, including "log" and "ln".

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First of all the second equation shows up as an unintelligible expression. So I will not be able to help you with that. But perhaps the solution to the first equation will help you figure out how to do the same for whatever the second equation is.

I assume by "equivalent equation" you mean "equivalent exponential equation". If I'm wrong then you'll have to re-post your question making it clearer.

In general, the logarithmic equation:

is equivalent to the exponential equation:


Using this pattern on:

we get:

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Expressions of the form:

are very simple to simplify. Your expression

is not quite in that form because the base of the exponent, 4, is different from the base of the logarithm, 2. But here is where we can figure out a relatively simple solution. Aren't 4 and 2 powers of each other? ( and ). So we should be able to convert one base, the 4 or the 2, into the other. I'm going to change the base of 4 into a base of 2. Replacing the 4 with we get:

At this point we have a power of a power. The rule for the exponents for this is to multiply the exponents:


We have the bases matching. But we are still not quite in the desired form. The 2 in front of the log is the problem. But fortunately there is a property of logarithms, , which allows us to move a coefficient in front of the log into the argument of the log as its exponent. Using this property we can move the 2 (which is in our way) into the argument as its exponent:

which simplifies to:

We now have the desired form. If you don't yet know how this simplifies, let's review what logarithms are. Logarithms in general are exponents. Base 2 logarithms are exponents one would put on a 2. More specifically,

is the exponent one would put on a 2 to get a result of 9. And look at where

is in our expression. It is the exponent on a 2! So our expression is "2 to the power that you would put on a 2 to get a result of 9". I hope it is clear that this means that
must be a 9!

If this is not clear then either:

• Memorize: ; or
• Learn to simplify it as follows. Let's say that:
  
  Find the base a log of each side:
  
  Use the property of logarithms mentioned earlier. This time, however, we are going to in the other direction: move the exponent of the argument out in front.
  
  We should know that so this simplifies to:
  
  The equation now says that two base a logs are equal. If so then the arguments must be equal. So:
  
  Since and :
  

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If

is the correct expression then please post it as:
(cube root of 2)((cube root of 4) subtract (2 times the cube root of 32))
The extra parentheses make it clear. Without them
cube root of 2(cube root of 4 subtract 2 times the cube root of 32)
could be interpreted as:


If

is not correct then you will have to re-post your question because I'm going to simplify this expression.

We could start with either using the Distributive Property to multiply by the or we could simplify since 8 is a factor of 32 and 8 is a perfect cube (). Looking ahead I can see that using the Distributive Property first will make the rest much easier. So that is how I will start:

Multiplying the roots together, using the property of radicals we get:

Simplifying inside the radicals we get:

8, as was mentioned earlier, is a perfect cube. And so is 64! . So both cube roots simplify:

Continuing to simplify:

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A general procedure for solving equations like this is:

1. Isolate a square root (that has a variable in it).
2. Square both sides of the equation.
3. If there is still a square root (that has a variable in it), then repeat steps 1-3.
4. At this point the equation should have no square roots (that have a variable in it). Use appropriate techniques to solve whatever kind of equation you now have.
5. Check your solutions. This is not optional! Whenever you square both sides of an equation, like step 2, you must check for what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. They can occur even if no mistakes have been made. You must check for them and reject them if you find any.

Let's see this in action:
1. Isolate a square root.
The third square root does not have x in it so that is not a square root to isolate. we should isolate one of the other two square roots. I'm going to add the second square root to each side and subtract 6 from each side giving us:


At this point we might be able to see that there is no solution to this equation. Here's the logic:

• Since the radicands (the expression within a radical is called a radicand) must not be negative we can tell that whatever x turns out to be x must be greater than or equal to 2.
• No matter what x turns out to be
• Since ,
• Also . This makes (IOW: negative)

Putting this altogether:
-
- is negative
- The equation says that a smaller square root plus a negative is equal to a larger square root.
- There are no solutions to the equation because it is impossible to take a smaller number , add a negative and get a larger square root . (Note: An exception to this is when both square roots are between 0 and 1). Since , . So we do not have this exception in this problem. So there are no solutions to this equation.

Note: I presented this logic because it is actually the "easy" way to solve this problem. But if you don't understand this or if you don't feel comfortable presenting this as your "work" or if your teacher would not accept such a logic-based solution, I will continue on with the procedure I outlined at the start...)

2. Square both sides.
Squaring the right side is as easy as it looks. But squaring the left side correctly will not be easy. It is not just squaring each term!! We have to multiply

which is done by multiplying each of the three terms in the first factor by each of the three terms of the other factor and then adding like terms, if any. Before we start multiplying I'm going to change the subtraction into the equivalent addition because it is easier to avoid errors with additions:


Squaring both sides:

Simplifying...

(Note: I'm not multiplying for a reason.) Adding like terms...


3. We still have a square root with a variable in the radicand so it's back to step 1.

1. Isolate a square root (that has a variable in it).
Subtracting x and 5 from each side:

Adding to each side:

Factoring out (which is why I didn't multiply ):

Dividing by :

(Note: Again we can save time and effort if we pay attention to what this equation says. The left side is a positive square root. The right side is a fraction with a positive numerator (since ) and a negative denominator (since ) making the whole fraction negative (positive divided by negative is negative.) Since the positive left side cannot be equal to the negative right side, there are no solutions. But we will forge ahead with the procedure...)

2. Square both sides.
Squaring the isolated square root will be easy, as always. To square the right side I am going to take advantage of the pattern to square the numerator and denominator:

Simplifying...




3. If there are square roots with a variable in it...
There are no more square roots with a variable. So on to step 4!

4. Solve the equation.
All we have to do is add 2:


5. Check your solution.
Use the original equation to check:

To make things a little easier I'm going to get a decimal approximation for , use that to get a decimal approximation for x and then use that for the x in the check. (Note: Because of the use of approximations we may not get perfectly equal sides of the equation. If we get almost equal sides then we can hope that the answer checked. The only way to be sure is to avoid the use of approximations, i.e. check with , not some decimal.)
, rounded to 3 places, is 1.732. Using this to get an approximation for x:






Now we can check:

Simplifying...



This is not close enough to equal to pass the check. (Again, if you want to be sure, don't use decimal approximations during the check!) So this solution must be rejected/discarded. And since we found no other possible solutions, there are no solutions to your original equation.

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Is "x2" or ? The problem is much easier if it is x*2. I'll solve it both ways, starting with :

On both sides of the equation, the variable is in the exponent of the argument of a log. There is a property of logarithms, , which allows us to move such exponents out in front of the log. Using this property on our equation we get:

On the left side we can use the Distributive Property to multiply:


With the term this is a quadratic equation. And with the log's in there, the answer is not going to work out to be a nice whole number (or even a fraction). So I am going to use a calculator to find all the logs (and round the log to the nearest 3 decimal places):


which simplifies to:

Subtracting the squared term from each side:

Then using the Quadratic Formula:

Simplifying:





which is short for:
or
which simplify as follows:
or
or
These are decimal approximations for the solutions to your equation if x2 meant

If x2 meant x*2 (or 2x)...
The steps are mostly the same so I'll leave out the commentary except to explain the differences.



This is not a quadratic equation. With this we want all the x terms on one side and the other terms on the other side. Subtracting the first term of the left side from both sides gives us:

Factoring out x from the terms on the right side:

And then dividing by :

This is an exact expression for the solution to the equation if x2 meant x*2 or 2x. Without having to deal with a quadratic equation made it much easier to avoid using our calculator (to get decimal approximations).

If you want or need decimal approximations for this solution, you can use the values for the logs in the first solution (when x2 meant and then simplify the left side.

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When I started thinking about how to simplify the expression one thing I noticed was that all the arguments (except 51 and 17) were powers of 3:





Then I noticed that the logs with arguments of 51 and 17 can be rewritten as a single log whose argument would also be a power of 3. All these powers of 3 will make simplifying the expression easier.

I'll start by rewriting the arguments (except 51 and 17) as powers of 3:

As we proceed we will be using several properties of logarithms:

1. 
2. 
3. 

The first one we will use is #2. The numerator with log(51) and log(17) matches the right side of that property. So, according to that property, we can rewrite the two logs as a single log of 51/17:

And since 51/17 = 3 we get:


To use properties #1 and #2 the bases of the logs must be equal and the coefficients of the logs must be 1's. We will continue to use these properties to combine two logs into a single log. So we want all the logs to have coefficients of 1. Several of the logs do not have coefficients of 1, specifically the last log in the denominator and the last two logs terms. Fortunately property #3 gives us a way to handle these coefficients. The right side of Property #3 shows that the log has a coefficient of "n". The left side shows us that we can move the coefficient, n, into the argument as its exponent. Using this property on the three logs which do not have a coefficient of 1 we get:

with the second-to-last log simplifying, according to the rules for exponents:

Now we can start using properties #1 and #2 to combine the remaining logs. Using property #2 on the first two logs:

Again we can use the proper rule for exponents, subtract, to simplify the new argument:

Next we can use property #2 again to combine the two logs in the denominator:

Again we subtract the exponents in the new argument:

Next we can use property #1 to combine the last two logs. (We use this property because its logs, like ours, have a "+" between them. All the previous pairs of logs had a "-" between them which is why we used property #2.)

This time the rule for the exponents in our new argument is to add the exponents:

Next we can use property #1 again to combine the first and last logs. (This expression is all addition so we are allowed to change the order of the terms!)

Again the rule for the exponents is to add:

Next I will use property #3 again. This time I am going to use it in the reverse direction. I'm going to move the exponent on the argument of the log in the denominator back out in front. (By doing this I can get the two logs in the fraction to match (will allow us to cancel them.)

Now the log(3)'s in the fraction cancel out:

leaving:

Last of all we can simplify our last remaining power of 3:

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Since you asked for general help and since radicals are used for all types of roots (square, cube, 4th, 5th, etc.) I can only provide very general help:

1. Determine which kind of root it is.
2. (Note: "Radicand" is the name for whatever expression is found inside a radical.) Try to find as many factors of the radicand that are powers which match the type of root as possible. For square roots look for perfect square factors, for cube roots look for factors that are perfect cubes, etc. (Note: if the only perfect square/cube/etc. factor you can find is 1, then the radical will not simplify and you can stop here.)
3. Rewrite the radicand in factored form. I like to put all the perfect square/cube/etc. factors in front and the other factor at the back.
4. Use a property of radicals, , to break up the radical so that each perfect square/cube/etc. factor is in its own "personal" radical. The other factors, if any can all be lumped into a single radical.
5. The radical(s) containing the perfect square/cube/etc. factors will all simplify.
6. If there were more than 1 perfect square/cube/etc. factors then simplify/multiply what is outside the radical.

Let's look at a couple of examples:
(assuming x and y are both positive)
Finding perfect square factors:

Reordering to put the perfect squares in front:

Separating the perfect squares into their own radical(s):

Each square root of a perfect square simplifies:

Simplify what's outside:



Finding perfect cube factors. (Note: and 8 is a factor of 72.)

Reordering to put the perfect cubes in front:

Separating the perfect cubes into their own radical(s):

Each cube root of a perfect cube simplifies:

Simplify what's outside:

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A nice equation with excellent use of parentheses to make it clear! But what are you/we supposed to do with it?? {Please re-post and include the instructions with the problem.

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refers to the positive number/expression that, when squared, results in 5x. But the equation says that is a negative number, -5. A positive square root cannot be equal to a negative number. IOW: There is no solution to this equation!

If you don't notice this and proceed to solve the equation, as you did, then you can still find out that there is no solution. Any time you square both sides of an equation, as you did, you must check your answers! Let's check your solution, x = 5, using the original equation:

Checking x = 5:



Since this is false, the check fails!! x=5 is not a solution. And since we must reject the only "solution" you found, there are no solutions to your equation.

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The error is that when the division was changed to a multiplication, the first fraction was inverted. It should be the second fraction (only) that gets inverted.


So we are good up to here. But with

we have inverted the wrong fraction. It should be:

After canceling we get:

which simplifies to just:
r+2