SOLUTION: 5^2log(base5)^x

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Question 624326: 5^2log(base5)^x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, I'm assuming your expression is:
5%5E%282log%285%2C+%28x%29%29%29
If this is correct, then the "x" in your expression is not an exponent, even though it kind of looks like one. The "x" is the argument to the base 5 logarithm. The confusing way you posted this may well be the reason it has taken this long for anyone to respond, You might want to click on the "Show Source" link above so you can see what I've typed to get these logs to display as they do here on algebra.com

A basic property of logarithms is:
a%5Elog%28a%2C+%28q%29%29+=+q

Your expression very nearly matches the pattern of the property. However the property pattern does not have a "2" in front of the log in the exponent like your expression does. So if we can find a way to get rid of the 2 we can use the property to simplify your expression.

One way to get rid of the 2 is to use another property of logarithms:
q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29
which allows us to move a number in front into the argument as its exponent:
5%5E%28log%285%2C+%28x%5E2%29%29%29
Now we can use the other property, with the "a" being "5" and the "q" being x%5E2. The property tells us that an expression matching the pattern of the left side is equal to the "q". So according to this property
5%5E%28log%285%2C+%28x%5E2%29%29%29
is equal to
x%5E2

Another way to "get rid of the the 2":
One property of exponents is:
a%5E%28p%2Aq%29+=+%28a%5Ep%29%5Eq
Usually we use this property to rewrite powers of a power. But as as often true, this property works in both directions. We can rewrite an exponent that is a product as a power of a power. So we can rewrite
5%5E%282log%285%2C+%28x%29%29%29
as
%285%5Elog%285%2C+%28x%29%29%29%5E2
Inside the parentheses matches the pattern of the a%5Elog%28a%2C+%28q%29%29+=+q property. So it is equal to x:
%28x%29%5E2
which simplifies to
x%5E2