SOLUTION: solve for x: log2x^3 - logx = log16 - logx

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Question 719524: solve for x: log2x^3 - logx = log16 - logx
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%282x%5E3%29%29+-+log%28%28x%29%29+=+log%28%2816%29%29+-+log%28%28x%29%29
First we'll use a property of logarithms, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29, to combine the two logs on each side of the equation:
log%28%28%282x%5E3%29%2Fx%29%29+=+log%28%2816%2Fx%29%29
The first fraction simplifies:
log%28%282x%5E2%29%29+=+log%28%2816%2Fx%29%29

We now have an equation that says that two base 10 logs are equal. In other words: The power of 10 that results in 2x%5E2 is equal to the power of 10 that results in 16/x. If you put a particular exponent on a 10 there is only one result. So if we put a particular exponent on 10 we should get the same result. This means that the results must be the same:
2x%5E2+=+16%2Fx
With the x out of the logarithms, we can now solve for it. Multiplying each side by x:
2x%5E3+=+16
Dividing by 2:
x%5E3+=+8
Cube root:
x+=+2

Last of all, we check. This is not optional when solving logarithmic equations like yours. You must at least check to see that all arguments and bases of all logarithms are valid when the variable is equal to the solution. (Valid arguments and bases are positive. Also, bases may not be 1.) If a "solution" makes any argument or base of any log invalid we must reject it.

Use the original equation to check:
log%28%282x%5E3%29%29+-+log%28%28x%29%29+=+log%28%2816%29%29+-+log%28%28x%29%29
Checking x = 2:

Simplifying:
log%28%282%288%29%29%29+-+log%28%282%29%29+=+log%28%2816%29%29+-+log%28%282%29%29
log%28%2816%29%29+-+log%28%282%29%29+=+log%28%2816%29%29+-+log%28%282%29%29
We can see that all arguments and bases are positive so this solution passes the check. (We can also see that 2 is the correct solution (i.e. we didn't make any mistakes) because the left side matches the right side.

P.S. In retrospect, we could have solved this a little faster if we had started by adding log(x) to each side of your original equation instead of using that property to combine the logs.