SOLUTION: Can you help me solve for x in the equation {{{(log(2, 25))(log(x, 16))=4}}}? Thanks!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you help me solve for x in the equation {{{(log(2, 25))(log(x, 16))=4}}}? Thanks!      Log On


   

Question 275956: Can you help me solve for x in the equation %28log%282%2C+25%29%29%28log%28x%2C+16%29%29=4? Thanks!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28log%282%2C+%2825%29%29%29%28log%28x%2C+%2816%29%29%29=4
Equations with logarithms are easier to work with and solve when the bases of the logarithms are the same. So we'll start by using the change of base formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to change the base of one of the logarithms. Changing the base of the base x logarithm to base 2 looks promising to me because 16 is a power of 2:
%28log%282%2C+%2825%29%29%29%28log%282%2C+%2816%29%29%2Flog%282%2C+%28x%29%29%29=4
Since 2%5E4+=+16, log%282%2C+%2816%29%29+=+4. Substituting this in we get:
%28log%282%2C+%2825%29%29%29%284%2Flog%282%2C+%28x%29%29%29=4
Multiplying we get:
%284log%282%2C+%2825%29%29%29%2Flog%282%2C+%28x%29%29=4
Multiplying both sides by log%282%2C+%28x%29%29 we get:
4log%282%2C+%2825%29%29=4log%282%2C+%28x%29%29
Dividing both sides by 4 we get:
log%282%2C+%2825%29%29=log%282%2C+%28x%29%29
So x must be 25!