SOLUTION: Log (base of square root 3) (square root of 9 with a 3 on the root in front) ^5x

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Question 238433: Log (base of square root 3) (square root of 9 with a 3 on the root in front) ^5x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28sqrt%283%29%2C+%28%28root%283%2C+9%29%29%5E%285x%29%29%29
If this is incorrect, please repost using parentheses to make things clearer. (BTW, root%283%2C+9%29 is called "the cube root of 9")

There is no equals sign. Is there supposed to be one? If yes, please repost with the entire equation. Without an equals sign this is just an expression and all you can do with an expression is simplify it.

All roots can be written as fractional exponents and rewriting the cube root of 9 as a fractional power of 9 will help us simplify:
log%28sqrt%283%29%2C+%28%289%5E%281%2F3%29%29%5E%285x%29%29%29
Using the property of exponents, %28a%5Ep%29%5Eq+=+a%5E%28%28p%2Aq%29%29, on this we get:
log%28sqrt%283%29%2C+%289%5E%28%285%2F3%29x%29%29%29
Now we can use the property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent in the argument out in front:
%285%2F3%29x%2Alog%28sqrt%283%29%2C+%289%29%29
And finally we need to figure out the logarithm: log%28sqrt%283%29%2C+%289%29%29. We could use changing the bases and then using calculators. Or we could just be clever. If you understand what logarithms represent, you know that log%28sqrt%283%29%2C+%289%29%29 represents the exponent for sqrt%283%29 that results in 9. So what is "the exponent for sqrt%283%29 that results in 9"? Well %28sqrt%283%29%29%5E2+=+3 and 3%5E2+=+9 So %28%28sqrt%283%29%29%5E2%29%5E2+=+%28sqrt%283%29%29%5E4+=+9. So "the exponent for sqrt%283%29 that results in 9" is 4. Substituting this in for log%28sqrt%283%29%2C+%289%29%29 we get:
%285%2F3%29x%2A4
which simplifies to
%2820%2F3%29x