SOLUTION: Can I please ask for help to solve x from this question? {{{log(x, (3)) = 4}}}

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Question 819231: Can I please ask for help to solve x from this question?
log%28x%2C+%283%29%29+=+4
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Rewrite it in exponential form:
3+=+x%5E4
4th root of each side:
0%2B-+root%284%2C+3%29+=+x
(Note: The zero is there only because algebra.com will not let me use the "plus or minus" sign without a number in front. You should not use the zero.)

Check. This is not optional. Use the original equation to check:
log%28x%2C+%283%29%29+=+4
Checking x+=+root%284%2C+3%29:
log%28root%284%2C+3%29%2C+%283%29%29+=+4
The base is positive but not 1. So it is valid. The argument is positive so it is valid, too. So this solution checks out.
Checking x+=+-root%284%2C+3%29:
log%28-root%284%2C+3%29%2C+%283%29%29+=+4
The base is negative. This is an invalid base. So we reject this "solution".

The only solution to your equation is: x+=+root%284%2C+3%29. (If you need this in the form of a decimal approximation, use 3%5E0.25 in your calculator.)