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Equations are easier if there are no fractions, I think you'll agree. So we'll start by eliminating the fractions. The asiest way to eliminate the fractions in an equation is to multiply both sides by the Lowest Common Denominator (LCD). And to find the LCD we need to factor the denominators:

The LCD is the product of all the different factors. So the LCD here is:
(x+2)x
This is what we will multiply both sides by:

On the left side we need to use the Distributive Property:

Now we can cancel:

leaving:

which simplifies as follows:

Without the fractions, this is a very simple equation to solve. It is a quadratic equation so we want one side equal to zero. So subtract 12 from each side:

Now we factor:
3x(x + 2) = 0
From the Zero Product Property we know that this product can be zero only if one of the factors is zero. So:
3x = 0 or x+2 = 0
Solving these we get:
x = 0 or x = -2

With equations where the variable is in one or more denominators, it is important to check your answers. We must make sure no denominators are zero! Always check with the original equation.

Checking x = 0:

which simplifies to:

As you can see, two of the denominators are zero. For this reason we must reject x = 0 as a solution. (If even only one denominator was zero we would still reject the solution.)

Checking x = -2:

which simplifies to:

Again we get zero denominators! We must reject this solution, too.

So there are no solutions to your equation!

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Your "equation" has two equals sigms. Please correct and repost.

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Equations are easier if there are no fractions, I think you'll agree. So we'll start by eliminating the fractions. The asiest way to eliminate the fractions in an equation is to multiply both sides by the Lowest Common Denominator (LCD). And to find the LCD we need to factor the denominators:

The LCD is the product of all the different factors. So the LCD here is:
(5)(x+1)(x-1)
This is what we will multiply both sides by:

On the left side we need to use the Distributive Property:

Now we can cancel:

leaving:
(x-1)(2) - (5)3 = 5(x+1)4
which simplifies to:
2x - 2 - 15 = 20(x+1)
2x - 17 = 20x +20
Without the fractions, this is a very simple equation to solve. Subtract 2x from each side:
-17 = 18x + 20
Subtract 20 from each side:
-37 = 18x
Divide both sides by 18:
-37/18 = x

With equations where the variable is in one or more denominators, it is important to check your answers. We must make sure no denominators are zero! A quick mental check and you can see that -37/18 does not make any of the denominators zero. So that is out solution.

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Since this fraction does not simplify. I'm guessing that the actual fraction is:


Simplifying fractions involves canceling factors that are common to the nnumerator and denominator. To cancel factors we need factors.

So we start by factoring. The numerator has a GCF of 2 which we can factor out. And the denominator is a difference of squares so it factors easily.

We don't have common factors yet so we keep factoring.

The second factor in the numerator

• doesn't fit any of the factoring patterns
• has too many terms for trinomial factoring
• doesn't appear to be factorable by grouping.

So it seem that we need to factor by trial and error of the possible rational roots. The possible rational roots are the ratios, positive and negative, which can be formed using a factor of the constant term (at the end, 18 in your case) over a factor of the leading coefficient, 4. The factors of 18 are 1, 2, 3, 6, 9 and 18. The factors of 4 are 1, 2 and 4. So there are a lot of possible rational roots. But we're only interested in factors that match a factor in the denominator. So the only roots worth trying here are 2 and -2. I'm going to try 2 first. To check a rational root is probably easiest with synthetic division:

2 |  4  -2  -3  -18
         8  12   18
    ---------------
     4  -6   9    0

The remainder is zero so 2 is a rational root and (x-2) is a factor. And, from the numbers in front of the remainder, the other factor is . So now our factored fraction is:

We could try to continue to factor the numerator but we can see that the rational roots of do not include -2. So x+2 will not be a factor. Since any other factors do not help we won't bother factoring any more. Now we can cancel the common factor of x-2:

which simplifies to:

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There are a variety of techniques you learn when factoring. One of these techniques is to factor by using certain patterns:

• Difference of squares:
• Sum of cubes:
• Difference of cubes:
• Perfect square trinomials:
  • 
  • 

This problem happens to be one where these patterns are all you need to factor it. Your expression has only two terms. This is too many terms for the last two patterns. And you expression has a "-" between the two terms so we can't use the second pattern (yet).

So the question is, is a difference of squares or a difference of cubes? Since exponent that is divisible by 2 is a perfect square and since any exponent that is divisible by 3 is a perfect square, and since 6 divisible by both 2 and 3, the answer to the question is both! So we can start with either pattern. I'll do it both ways.

Difference of squares first:

Using the pattern we get:

When factoring you keep going until you cannot factor any further. The two factors we have now are clearly a sum of cubes and a difference of cubes, respectively. We can use those patterns on them:


Using difference of cubes first:

Using the pattern we get:

The last factor can be simplified and the first factor is a difference of squares we can factor:

The first two factors match two of the factors we found earlier. Earlier, however, we ended up with four factors and here we only have three. So somehow
the third factor must factor in the other two factors we found earlier: and . But the factoring involved in turning into is very complex and I will not do it here. I'll just say then, when faced with choosing between difference of squares and difference of cubes, choose difference of squares first. It makes things easier.
*******************
In response to the note you included in your thank you...
The answer you mention is the same as the one above where we started by using the difference of cubes first. (I changed it so it shows up in red so you can find it more easily.)

But it remains true that this factored expression is not as fully factored as the one you get when you start with the difference of squares. (I changed this one to show up in green.) The red is correct. The green is correct and a better answer.

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So square the 2n:

Now it is in the form ax^2 + bx + c

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To find zeros we need to factor the polynomial. The straightforward way to factor this would be to simplify it first. But this is a lot of work. We would have to square 2x+7 and both square and cube x-1, add the like terms and then try to factor what we end up with.

But you may notice that there are already common factors on each side of the subtraction in the middle. We can use this to our advantage and make the problem much easier. First I am going to rewrite the expression to make what I do clearer:

The GCF of the expressions on each side of the subtraction are in red. I will now factor out this GCF from each side:

Now I'll simply expression in the last factor:



And we now have p(x) factored. And the zeros will be the values of x that make the factors zero:
2x+7 = 0 or x-1 = 0 or 5x+22 = 0
Solving these we get:
x = -7/2 or x = 1 or x = -22/5
These are the zeros of p(x).

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"Exactly divisible" means that there will be no remainder if you divide. So let's divide and see what "a" and "b" have to be in order for there to be no remainder:

        2x   + 4
        _______________________________
x^2-1  /2x^3 + 4x^2 + 2ax      + b
        2x^3        - 2x
        ---------------------
               4x^2 + 2ax + 2x + b
               4x^2            - 4
               -------------------
                      2ax + 2x + b + 4

Since we're supposed to have no remainder, the expression at the bottom must work out to be zero. And the expression will be zero if the like terms combine to make zeros. So

2ax + 2x = 0     and  b + 4 = 0
2x(a + 1) = 0         b     = -4
a = -1

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When multiplying factors with exponents and the same base, like yours, the rule is to add the exponents. It does not matter what kinds of numbers the exponents are. just add them:

As always, when you add fractions you must have the same denominator:

which gives us:

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Each quadrant represents a certain combination of values for x and y:

                  x        y
Quadrant I     Positive  Positive
Quadrant II    Negative  Positive
Quadrant III   Negative  Negative
Quadrant IV    Positive  Negative

So if we can figure out what values x can be and what values y can be, then we can figure out what what quadrants the graph will be in.

In your equation x is the exponent (of 10). What kinds of numbers can exponents be? Answer: Exponents can be any number (positive, negative, zero, whole numbers, fractions, etc.). So x can be any number.

In your equation the y is . What kinds of numbers can powers of 10 be? With some careful thought we should be able to determine that powers of 10 can never be zero or negative. Remember:

• A zero exponent of 10 results in 1, not zero:
• Negative exponents of 10 do not result in negative answers. For example,

So we have found that x can be anything but y has to be positive. In which quadrants are positive y's found? The answer to this question is the answer to your problem.

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log(x-9) = 1 - log(x)
Solving an equation where the variable is in the argument (or base) of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has that "non-logarithmic" term (the 1), reaching the second form (which is all logarithms) will not be easy. So we will aim for the first form. We need the logarithms on one side so we'll start by adding log(x) to each side:
log(x-9) + log(x) = 1
Now we need to combine the logarithms into one. These are not like terms so we cannot add them. But there is a property of logarithms, , which allows us to combine two logarithms with a "+" between them (as long as the bases are the same and the coefficients are 1's). Your bases are the same, 10, and the coefficients are 1's so we can use the property:
log((x-9)*(x)) = 1
which simplifies to:

We now have the equation in the first form. With this form we proceed by rewriting the equation in exponential form:

which simplifies to:

The variable is now out of the logarithms. This is a quadratic equation so we want one side equal to zero. So we'll subtract 10 from each side:

Now we factor it (or use the Quadratic Formula). This factors easily:
(x-10)(x-1) = 0
From the Zero Product Property we know that this product is zero only if one fo the factors is zero. So:
x-10 = 0 or x-1 = 0
Solving each of these we get:
x = 10 or x = 1

When solving logarithmic equations, it is important (not just a good idea) to check your answers. Always use the original equation to check:
log(x-9) = 1 - log(x)
Checking x = 10:
log(10-9) = 1 - log(10)
log(1) = 1 - log(10)
Since log(1) = 0 and log(10) = 1 this simplifies to:
0 = 1 - 1
0 = 0 Check!

Checking x = 1:
log(1-9) = 1 - log(1)
log(-8) = 1 - log(1)
At this point we have a problem. The first logarithm has a negative argument. Arguments of logarithms can never be negative or zero. So we must reject this solution.

That makes x = 10 the only solution to your equation.

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Solving equations where the variable is in an exponent usually involves using logarithms. If you want an exact answer, then you can use a logarithm of any base. But the "best" base to choose the same as the base for which the variable is part of the exponent. In this case this would be base 7. But if you want a decimal approximation then you should choose a base your calculator "knows" (like base 10 or base e (ln)). I'll do the problem both ways.

Using base 7 logarithms to get the simplest exact answer:

Now we use a property of logarithms, , to move the exponent out in front. This property is precisely the reason we use logarithms on problems like this. It allows us to move the variable out of the exponent where we can "get at it". Using this property on the first logarithm we get:

By definition, . (This is why we chose base 7 logarithms.) So the equation simplifies to:

Now we just subtract 1 from each side:

This is an exact expression for the solution to your equation.

Using base 10 (or base e) logarithms (because we want a decimal approximation):

Using the property to move the exponent out in front:

Dividing both sides by log(7):

Subtracting 1 from each side:

Now we can get out our calculators and

1. Find the two logarithms.
2. Divide the two logarithms
3. Subtract 1 from the result of the division.

This should result in something very close to the 0.933.

P.S. If you find the exact answer then decide that you want a decimal, then you can take

and use the base conversion formula, , to convert the base 7 logarithm into base 10 (or base e) logarithms.

P.P.S. If you use base e instead of base 10 logarithms to find a decimal approximation, you end up with the same answer! (Try it an see.)

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As it is, this logarithm cannot be found. No calculator "knows" base 8 logarithms and 32 is not a known power of 8. So what we need to do is transform this logarithm into an expression we can find. Fortunately there is a base conversion formula, , which we can use to change this base 8 logarithm into a logarithm we can find.

So we need to choose a base to change the base 8 logarithm into. I'll choose two different bases.
Base 2. Since 8 is a power of 2 (), using base 2 has some promise of making the problem easier. Using the conversion formula to convert the base 8 logarithm into an expression of base 2 logarithms we get:

As stated earlier 8 is 2 cubed. So the logarithm in the denominator is 3. And, as it turns out 32 is also a power of 2! . So the logarithm in the numerator is 5. This makes our expression:
5/3
And we are finished (except perhaps changing the improper fraction into a mixed number).

If 8 had not been a power of 2 or if we didn't recognize that 8 (or 32) was a power of 2, then we would need to cahnge the base to one our calculator "knows" (like base 10 or base e (ln)). Converting your expression into base 10 logarithms we get:

And now we would use our calculators to find the two logarithms and them divide. The first way is better than this way. For one, calculators use decimal approximations for most logarithms. Secondly, the decimal answer you get from a calculator may by hard to recognize as equal to the fraction 5/3.

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When working with fractions you know that your answer should be reduced and if the fraction is improper it should be converted to a mixed number. When working with square roots:

• The square roots should be in "reduced" form. This means that there should be no perfect square factors in the radicand (the expression within the radical).
• Denominators must be rational. This means:
  • No fractions within a radical
  • No square roots in any denominators.

Your square roots have no perfect square factors (yet). But you do have square roots in a denominator so we need to rationalize it.

Your denominator has two terms and rationalizing two-term denominators is trickier that you might think. You might think to square the numerator and denominator. But there are two things wrong with that:

1. Squaring the numerator and denominator is not a valid operation. The new fraction you get will not be the same as the original equation.
2. Even if squaring the numerator and denominator was allowed, it would still not rationalize the denominator. This is true because is NOT equal to ! Exponents do NOT distribute. means and if you multiply it out properly (using FOIL or otherwise), you get .

So how do we rationalize two-term denominators? Well I hope the expression provides a hint. It should look familiar. There is a pattern used in factoring that involves an expression like this:

This points us toward an answer. It tells us that we can take a two-term expression and turn it into a two-term expression of perfect squares. If we have a two-term expression with a "+" between them, like (p+q), we can get perfect squares if we multiply it by (p-q). And if we have a two-term expression with a "-" between them, like (p-q) and like your denominator, we can get perfect squares by multiplying by (p+q).

So to rationalize your denominator we need to multiply it by . And if we multiply the denominator we have also have to multiply the numerator by the same thing:

The multiplication in the denominator is easy. The pattern tells us what we get. In the numerator we can use FOIL, the pattern for or other proper techniques:

which simplifies to:

There are no square roots in the denominator and the only square root remaining has no perfect square factors in its radicand. And the fraction will not reduce so we are finished.

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We want the equation in the form:
log(expression) = other-expression
So somehow we need to combine the two logarithms into one. These two logarithms are not like terms so we cannot subtract them. But there is a property of logarithms, , which can be used to combine two logarithms if all of the following are true:

• there is a "-" between them
• the bases of the logarithms are the same
• the coefficients of the logarithms are 1's

Your logarithms meet the first two but not the last. So now our goal is to get rid of the 3 in front of the second log. And fortunately there is another property of logarithms, , which can be used to move a coefficient of a logarithm into its argument as an exponent. Using this on your second log we get:

which simplifies to:

These are still not like terms so we still cannot subtract them. But we can now use the other property to combine them:

We now have the desired form. Once we have this form the next step is to rewrite the equation in exponential form:

which simplifies to:

Now the variable is out of the argument where we can "get at it". Solving this for x we start by multiplying both sides by 64 to get rid of the fraction:
3x + 10 = 1600
Subtracting 10 from each side:
3x = 1590
Dividing by 3 we get:
x = 530

When solving logarithmic equations, it is important (not just a good idea) to check your answers. Even if we've done everything correct so far, we need to make sure that each answer makes the argument of any logarithms positive.

Always use the original equation to check:

Checking x = 530:

which simplifies to

Both arguments are positive so it looks good. (You're welcome to finish the check on your own.)

The solution to your equation, then, is x = 530}}}

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We want the equation in the form:
log(expression) = other-expression
So somehow we need to combine the two logarithms into one. These two logarithms are not like terms so we cannot subtract them. But there is a property of logarithms, , which can be used to combine two logarithms if al of the following are true:

• there is a "-" between them
• the bases of the logarithms are the same
• the coefficients of the logarithms are 1's

Your logarithms meet the first two but not the last. So now our goal is to get rid of the 2 in front of the first log. And fortunately there is another property of logarithms, , which can be used to move a coefficient of a logarithm into its argument as an exponent. Using this on your first log we get:

These are still not like terms so we still cannot subtract them. But we can now use the other property to combine them:

We now have the desired form. Once we have this form the next step is to rewrite the equation in exponential form:

which simplifies to:

Now the variable is out of the argument where we can "get at it". Solving this for x we start by multiplying both sides by 5 to get rid of the fraction:

So
or
Simplifying these square roots we get:
or

When solving logarithmic equations, it is important (not just a good idea) to check your answers. Even if we've done everything correct so far, we need to make sure that each answer makes the argument of any logarithms positive.

Always use the original equation to check:

Checking

Both arguments are positive so it looks good. (You're welcome to finish the check on your own.)

Checking

As you can see the argument to the first logarithm is negative. We cannot allow this to occur. So we reject this solution.

The only solution to your equation, then, is

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When simplifying expressions of square roots there are two issues to address:

• Simplifying the square roots by extracting perfect square factors, if any
• Making sure any denominators are rational (i.e. no square roots in denominators.

To accomplish these tasks we often need one or both of the following properties of radicals:

• 
• 

These properties can and often are used in both "directions". For example the first property lets us change
into
It also lets us change
into

As you probably know already, there are often many ways to solve a problem. This is especially true with problems like this. So there are many ways to reach the simplified expression for this problem. Here's one way:
Use the second property to split the square roots of the fractions into fractions of square roots:

Now we'll multiply the two fractions using the first property to multiply the square roots:

Next I'll address the issue of rational denominators. We have a square root in the denominator which we cannot have in our final expression. To rationalize a one-term denominator like this we will multiply the numerator and the denominator by whatever square root that will make the denominator turn into a perfect square. The most obvious choice is to multiply by . And this will work. But the best choice is the one that creates the "smallest" perfect square. (It's kind of like adding fractions. You can add 1/2 and 1/4 if you turn both denominators into 8's. But it is better to use denominators of 4.) The best choice is to multiply by :

which results in:

The denominator simplifies to:

And the denominator is rational. Next we have to see if the remaining square root can be simplified. For this we look for perfect square factors. We should recognize, that is both a perfect square and it is a factor of the radicand ("radicand" is a word for the expression within a radical).

Now that we have identified a perfect square factor we factor it out:

Now we use the first property of radicals to split out the perfect square factor into its own square root:

And finally simplify the square root of the perfect square:

We have a rational denominator and there are no square roots left which have perfect square factors. This means we are finished.

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To find an inverse:

1. Swap the x's and y's. Wherever you see an x, write y and wherever you see y, write x. This creates the inverse equation.
2. If possible solve the new equation for y

Let's see how this works on your equation.
1. Swap the x's and y's:

2. Solve for y, if possible.
To solve our equation for y we start by rewriting the equation in exponential form. (This gets y out the of the argument of the logarithm.)

At this point we get stuck. We cannot solve this for y. (At least not for a single-valued expression for y.) So the inverse of your original function is not a function. The inverse can be expressed with any of the following:


( or )

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Solving a logarithmic equation often starts by transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has nothing but logarithms, we'll aim for the second form. This will require that we combine the two logarithms on the left into one. They are not like terms so we cannot use regular addition. But there is a property of logarithms, , which allows you to combine two logarithms of the same base with a "+" between them. Using this on the two logarithms on the left we get:
ln((x-6)*(x+1)) = ln(x-15)
We now have the desired form. With this form we have two logarithms that are equal. And if the logarithms are equal then the arguments must be equal:
(x-6)*(x+1) = x-15
Now we have an equation we can solve. Multiplying out the left side we get:

This is a quadratic equation so we want one side to be zero. Subtracting x from and adding 15 to each side we get:

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

From the Zero Product Property we know that this product is zero only if one of the factors is zero. Since both factors are the same, x-3 must be zero. Solving for x we get
x = 3

When solving logarithmic equations it is important (not just a good idea) to check your answers. And use the original equation when checking:
ln(x-6)+ln(x+1)=ln(x-15)
Checking x = 3:
ln((3)-6)+ln((3)+1)=ln((3)-15)
which simplifies to:
ln(-3)+ln(4)=ln(-12)
As you can see, two of the arguments are negative. Arguments of logarithms must always be positive. They can never be zero or negative. Values of x make an argument zero or negative are not allowed to be in the domain. So x=3 is not in the domain. We must reject this solution.

Since x=3 was the only possible solution and since we had to reject it, there is no solution for your equation. (IOW, the original equation was impossible to begin with.)

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Solving an equation with the variable in an exponent usually involves use of logarithms. Although any base of logarithm can be used, it makes things simpler if you use the same base of logarithm as the base(s) that have an exponent with a variable. Since both bases with exponents are e, then we'll use base e (ln) logarithms:

The reason we use logarithms on problems like this is this property of logarithms: . It allows one to take an exponent of the argument of a logarithm and move it in front of the logarithm. This is how we get the variable out of the exponent.

But to use this property, the exponent has to be the exponent of the entire argument. And our exponents apply to just the e's in the arguments and not to the 15 and the 13. Somehow we need to separate the 15 and the 13 from the e's. Fortunately there is another property of logarithms, , which can be used to separate factors of an argument into separate logarithms. Using this on our two logarithms we get:

Now we can use the other property to move the exponents (of the logs that have exponents) out in front:
ln(15) + (0.08t)ln(e) = ln(13) + (0.09t)ln(e)
And since ln(e) = 1 by definition this simplifies to:
ln(15) + 0.08t = ln(13) + 0.09t
Now that the variable is out of the exponents, we can solve for it. Subtracting 0.08t from each side we get:
ln(15) = ln(13) + 0.01t
Subtracting ln(13) from each side:
ln(15) - ln(13) = 0.01t
The two logarithms on the left cannot be subtracted. They are not like terms. But we can use yet another property of logarithms, , to combine them:
ln(15/13) = 0.01t
And finally we multiply both sides by 100:
100ln(15/13) = t
This is an exact expression for t.

P.S. When a problem refers to an exact solution it means "without a calculator" (because calculators use decimal approximations for more logarithms).

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What do the underscores, "_", mean?

And is the equation

or
?

Please repost your question making it clear. If needed use more English and less algebraic notation. For example the second equation would be:
base 2 log of x + base 4 log of x = 6
*******
Now that I know what the equation is...

To solve equations where the variable is in the argument(s) of logarithms, you usually start by transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Your equation, with the "non-logarithmic" term of 6 on the right, makes the second form more difficult to achieve. So we will aim for the first form.

The first form requires that one side is a single logarithm. So somehow we need to combine your two logarithms into one. Your two logarithms are not like terms so they cannot be added. In addition, a property of logarithms, , can be used to combine two logarithms which have a "+" between them. But this property requires that the bases of the two logarithms be the same. And your bases are different so we cannot use this property (yet).

So we need the bases the same. Fortunately there is a base conversion for logarithms, , which can be used to convert a logarithm of one base, "a", into an expression of another base, "b". We will use this to convert your base 4 logarithm into base 2:

And the denominator is a logarithm we can do "by hand". Since , :

or

These two logarithms are like terms so we can go ahead and add them:

The only thing left to do in order to achieve the desired form is the get rid of the 3/2. We can accomplish this by multiplying both sides by 2/3:

We finally have the desired form. With this form the next step to rewrite the equation in exponential form:

which simplifies to

And we have the answer.

Checking the answer is important (not just a good idea) with logarithmic equations. Always check using the original equation:

Checking x = 16:

Since and , and . This gives us:
Check.

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I'm going to assume that the equation is

because I don't see how to solve . If I am correct, then in the future please put parentheses around multiple term exponents, numerators, denominators and arguments to functions.

With variables in exponents we usually use logarithms to solve the equation. The base of the logarithm can be anything but it is often to your advantage to choose a base for the logarithm that matches the base for the exponent with the variable in it.

Your equation has two exponents with a variable in them so we have a choice of base e (ln) or base 6 logarithms. The problem says to solve for the exact value of x. This is code for "Don't use your calculator" because calculators provide only decimal approximations for most logarithms. So there is little advantage to one base over the other. (If decimal approximations were desired, then the base e logarithm would be the best choice because many calculators "do" base e (ln) logarithms and no calculators do base 6 logarithms.) I'll do the problem using both logarithms so you can see both ways.

Find the base e logarithm of each side:

On the left side we can use the property of logarithms, , to move the exponent in the argument out in front. On the right side we cannot do the same thing (yet) because the exponent in the argument applies just to the 6, not to the entire argument.


On the right side we need to separate the 8 and the in the argument. To our rescue comes another property of logarithms: . Using this on the right side we get:

Now we can use the other property to move the exponent out in front:
(x+3)ln(e) = ln(8) + x*ln(6)
(Note: The property that we just used to move the exponent out in front is exactly the reason we use logarithms on problems with variable in the exponents! It allows us to move the variables out of the exponent.)

The ln(e) is 1 by definition so the equation simplifies to:
x+3 = ln(8) + x*ln(6)
Now that the x's are out of exponents, all that is left is to solve for x. For this we need to gather the terms with x on one side and the terms without x on the other. Subtracting x and subtracting ln(8) from each side we get:
3 - ln(8) = x*ln(6) - x
On the side with the x terms we need to factor out x:
3 - ln(8) = x*(ln(6) - 1)
And finally divide both sides by (ln(6) - 1):

which is an exact solution for x. (Note: You cannot cancel the 3 in ln(6) with the 3 in the numerator! Nor can you cancel the 2 in ln(6) with the 2 in ln(8)! This fraction is a simple as it can get.)

Solving the problem with base 6 logarithms instead of base e (without the commentary):








Although this looks very different from our base e solution, it is just as exact, just as correct.

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Just use FOIL or whatever technique you've learned to multiply something like
(a + b)(c - d)

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Solving equations where the variable is in an exponent usually involves use of logarithms. But before we do that, it will make things simpler if we get rid of the 5 first. So we'll start by dividing both sides by 5:

Now we'll use logarithms. It doesn't really matter which base of logarithm we use but it makes things simpler if you use the same base for the logarithm as the base that has an exponent. So we will use base e (ln) logarithms:

Now we use a property of logarithms, , to move the exponent of the argument out in front of the logarithm. This gets the variable out of the exponent! In fact it is this property that is the reason we use logarithms on problems like this: to get the variable out of the exponent.

By definition, ln(e) = 1 so this simplifies to:

Now we have just one thing left. Divide both sides by 0.045:

This is the exact answer for x. If you want a decimal approximation, get out your calculator and calculate the expression on the left.

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There are two main issues when simplifying square roots:

• Finding and extracting prefect square factors, if any
• Making any denominators rational (i.e. No irrational numbers like square roots in denominators.)

These issues can be addressed in any order. Generally I like to address the rational denominators, first. Sometimes, if you extract the perfect square factors first and then rationalize the denominators, you end up having to extract perfect square factors again. (When you reduce fractions, you keep reducing until you can't reduce them any further. Extracting perfect square factors is like reducing fractions in the sense that you keeping going until you can't go any further.)

So I'll start with rationalizing the denominator. With a monomial (one-term expression) like this you rationalize the denominator by multiplying the numerator and denominator by whatever it takes to turn the denominator into a perfect square. With a denominator of the obvious choice is to multiply by . And it would not be wrong to do this. But there is a better choice. The smallest achievable perfect square is best. And with this denominator we can multiply by and get a perfect square. So this is what I will use. (If you use you will still get the answer right but it there will more work to do. It's a little like the using a denominator of 8 when adding 1/2 and 1/4.)

Multiplying we get:

and the denominator simplifies:

and the square root in the denominator is gone. Now we turn to extracting perfect square factors, if any, from the remaining square root:


Replacing the square roots of the perfect squares we get:

The x in the numerator cancels with one of the three factors of x in the denominator leaving:

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If the problem is as you posted it, then there is nothing you can do to simplify it.

However, if the problem is actually:

then there is something we can do. And it helps if you understand what a logarithm represents. Logarithms are exponents. In general

represents the exponent for "a" which results in "b". With your equation,
ln(x)
represents the exponent for e which results in x. And where so we find ln(x)? Answer: As an exponent on e! So

by the very definition of logarithms! So your equation simplifies to:
y = 2x

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Correct.

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Simplifying square roots is a matter of finding factors that are perfect squares.


Are there perfect square factors in 7825? One that should be obvious is 25. Factoring out 25 we get:

Using the property of radicals, we can separate the two factors into their own square roots:

And we can replace the first square root with its value of 5:

(You can check but I don't think 313 has any perfect square factors. If this is true then twe are done.


Again we look for perfect square factors. Since the last two digits are divisible by 4 then the whole number is divisible by 4 (and 4 is a perfect square).

Since the digits of 4131 add up to a number which is divisible by 9, then the entire number is divisible by 9 (which is a perfect square):

The digits of 459 also add up to a number divisible by 9:

51 has not perfect square factors so we are finished with the factoring. Now we split up the square root:




I'll leave the last one for you.

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To solve for x means to transform the equation so that it has the form
x = something
or
something = x

With your equation, to get the x by itself on one side of the equation, we will need to get rid of the 4, the 2 the -7 and the 5 (not in that order). To get rid of all these we need to "peel away" from the outside. So we start by getting rid of the 4 which is separate from the expression in parentheses and its exponent. We can divide both sides by 4 giving:

Still peeling away from the out side, the 5 is the next to go. We can find the 5th root of each side:

The left side simplifies to:

The next to go is the -7 (since it is separate from the term with x in it). We can just add 7 to each side:

(I put the 7 in front on the right side so that it cannot be thought of as part of the 5th root.) The 2 is the last to go. Divide both sides by 2:

which is the answer to the problem. (If you need a decimal approximation then use your calculator on the right side. To find a 5th root with your calculator, just raise 3 to the 0.2 power (since 5th roots are exponents of 1/5 and 1/5 as a decimal is 0.2).)

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In simplifying expressions with square roots you

• Reduce the square roots by factoring out perfect squares, if any
• Make sure the denominators are rational (i.e. leave no square roots in denominators).

The order in which these are done is not important. But you can save time by examining your expression and seeing if there are advantages to one way or another.

In your expression, we could start with simplifying the square root in the denominator (because 4, which is a perfect square, is a factor in 8. But after we're done with that we'd still have a square root in the denominator. This isn't bad but it means we're not saving any time by simplifying the square root in the denominator first.

Let's look at rationalizing the denominator. To do so we will multiply the numerator and denominator by any expression that makes the denominator a perfect square. The first possibility that comes to mind is to multiply . But there is a simpler possibility: . Multiplying the numerator and denominator by this will not only rationalize the denominator but it will also leave a square root in the numerator that cannot be simplified. In other words, this one step will accomplish everything that needs to be done:





Remember, this was just the quickest way to the answer. Multiplying by works, too, but you have to simplify the numerator afterwards. And we could start by simplifying the denominator before rationalizing it. But no matter what path you take to this problem, the answer above is the only fully simplified answer with a rational denominator.

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You're given A so:

We're solving for V so we need to get rid of the 6 in front and the exponent of 2/3. We can eliminate the 6 by dividing both sides by 6:

Now how do we get rid of the exponent? Well, to be perfectly correct, we are not getting rid of the exponent we are changing it to a 1 (since . So how do we change the exponent to a 1?

One way is to raise each side to the 3/2 power. (If you don't see why this works yet, you will shortly.)

On the right side, the rule is to multiply the exponents. 2/3 and 3/2 are reciprocals of each other. And what do you get when you multiply reciprocals? Answer 1! This is why I raised both sides to the 3/2 power:

• I knew that raising a power to a power would result in multiplying exponents
• I knew that I wanted an exponent of 1
• I knew that multiplying reciprocals always results in 1
• 3/2 is the reciprocal of 2/3

Now we only have to simplify the left side. If you are not yet comfortable with fractional exponents, it can be helpful to look at the exponent in factored form. The factor of three in the exponent says we'll be cubing something and the factor of 1/2 in the exponent says there will be a square root. And, since multiplication is Commutative, we can do these things in any order we choose. Since 2 is not a prefect square, doing the square root first has little appeal. So I'll start by cubing 2. 2 cubed is 8 and then we apply the square root giving us . So:

The only thing left is to simplify the square root. There is a perfect square factor in 8: