SOLUTION: How do you solve log4+log8=3logx-log2

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Question 417146: How do you solve log4+log8=3logx-log2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log(4) + log(8) = 3log(x) - log(2)
Solving equations where the variable is in the argument of a logarithm, like this equation, usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since all the terms in your equation are logarithmic terms, the second form will be easier to reach. All we have to do is find a way to combine the two terms on each side into one term.

There are no like terms on either side of the equation so we cannot just add or subtract to combine them. (Like logarithmic terms have logarithms of the same bases and same arguments.)

There are two properties of logarithms which allow you to combine two into one:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29

These properties require logarithms of the same base with coefficients of 1. The two logs on the left meet these requirements. On the right side the first log has a coefficient of 3. Before we can use one of the properties on the right side we will need to get rid of that 3 somehow.

Fortunately there is another property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29, which allows us to move a coefficient into the argument as its exponent. So we start by using this property to move the 3 into the argument on the left side:
log%28%284%29%29+%2B+log%28%288%29%29+=+log%28%28x%5E3%29%29+-+log%28%282%29%29
Now we can use the first two properties to combine the terms on each side. We use the first property on the left side (because of the "+" between the terms) and the second property on the right side (because of the "-" between them):
log%28%284%2A8%29%29+=+log%28%28x%5E3%2F2%29%29
which simplifies to:
log%28%2832%29%29+=+log%28%28x%5E3%2F2%29%29
We now have the second form. The next step with the second form is based on some basic logic. The equation says that two base 10 logarithms are equal. The only way this can be true is if the arguments are equal, too. So:
32+=+x%5E3%2F2
This equation is easily solved. Multiply both sides by 2:
64+=+x%5E3
And find the cube root of each side:
root%283%2C+64%29+=+root%283%2C+x%5E3%29
Since 64+=+4%5E3, the cube root of 64 is easy:
4 = x

Answers to equations like yours must be checked. You must ensure the answers make all arguments to all logarithms positive. Any "solution" that makes an argument zero or negative must be rejected (because arguments to all logarithms must be positive). These rejected solutions can happen any time and when they do happen it is not a sign that a mistake was made! So even expert mathematicians must check their answers on these equations.

Use the original equation to check:
log(4) + log(8) = 3log(x) - log(2)
Checking x = 4:
log(4) + log(8) = 3log((4)) - log(2)
We can instantly see that all four arguments are positive. So there is no reason to reject this solution. This is the required part of the check. Finishing the check is optional. It will show if we made a mistake. You are welcome to finish the check.

So the only solution to your equation is x = 4.