SOLUTION: Can you help me simplify {{{(125^(2log(5, x)))(3^(log(9, (x^10))))}}}? I believe it's {{{x^21}}}, but I'm not sure. Thanks!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you help me simplify {{{(125^(2log(5, x)))(3^(log(9, (x^10))))}}}? I believe it's {{{x^21}}}, but I'm not sure. Thanks!      Log On


   

Question 275957: Can you help me simplify %28125%5E%282log%285%2C+x%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29? I believe it's x%5E21, but I'm not sure. Thanks!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28125%5E%282log%285%2C+x%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29
The key to this problem is understanding that by definition:
a%5Elog%28a%2C+%28p%29%29+=+p
So we want to make the exponent of 125 a base 125 logarithm and to make the exponent of 3 a base 3 logarithm. There is a formula for changing bases, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, but we can't use it if the logarithm has a (visible) coefficient. So we need to start by using a 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 of the logarithm as its exponent. We will use this to move the 2 into the argument of the logarithm:
%28125%5E%28log%285%2C+%28x%5E2%29%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29
Now we can use the change of base formula to change the two exponents into base 125 and base 3 logarithms, respectively:

Since the cube root of 125 is 5 ans since cube root is an exponent of 1/3, 125%5E%281%2F3%29+=+5 and log%28125%2C+%285%29%29+=+1%2F3. And log%283%2C+%289%29%29+=+2. Substituting both of these into the denominators of the exponents we get:

Changing the divisions in the exponents into multiplying by the reciprocals we get:

Using our property to move coefficients into arguments again we get:

which simplifies to:
%28125%5E%28log%28125%2C+%28x%5E6%29%29%29%29%283%5E%28log%283%2C+%28x%5E5%29%29%29%29
By definition of logarithms
125%5E%28log%28125%2C+%28x%5E6%29%29%29+=+x%5E6 and 3%5E%28log%283%2C+%28x%5E5%29%29%29+=+x%5E5. Substituting we get:
%28x%5E6%29%2A%28x%5E5%29
which simplifies to:
x%5E11