SOLUTION: Hello, this is a logarithmic problem. I need a step by step explanation of: 16^log4(8). I know the answer is 64, but I need to show my work. Is it possible to solve this problem

Algebra ->  Equations -> SOLUTION: Hello, this is a logarithmic problem. I need a step by step explanation of: 16^log4(8). I know the answer is 64, but I need to show my work. Is it possible to solve this problem       Log On


   

Question 1148576: Hello, this is a logarithmic problem. I need a step by step explanation of:
16^log4(8). I know the answer is 64, but I need to show my work.
Is it possible to solve this problem by rewriting 16 as 2^4 and log 4 as 2^2 and 8 as 2^3. If so, please explain
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


There are many variations of how to simplify this expression. The method you are asking about is one of them. Here is how it might go....

16%5Elog%284%2C8%29

Rewriting each number as a power of 2....

%282%5E4%29%5E%28log%282%5E2%2C2%5E3%29%29

Now in the exponent %28log%282%5E2%2C2%5E3%29%29 you can use the "change of base formula"

log%28b%2Ca%29=+log%28n%2Ca%29%2Flog%28n%2Cb%29

and choose n=2 as the base:

%28log%282%2C2%5E3%29%2F%28log%282%2C2%5E2%29%29%29+=+3%2F2

So now the expression is

%282%5E4%29%5E%283%2F2%29+=+2%5E%284%2A%283%2F2%29%29+=+2%5E6+=+64

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Hello, this is a logarithmic problem. I need a step by step explanation of:
16^log4(8). I know the answer is 64, but I need to show my work.
Is it possible to solve this problem by rewriting 16 as 2^4 and log 4 as 2^2 and 8 as 2^3. If so, please explain
It is longer, but yes, it can be done!

16%5Elog+%284%2C+%288%29%29

log+%284%2C+%288%29%29 can be written as: log+%282%5E2%2C+%282%5E3%29%29, which means that: matrix%281%2C3%2C+%282%5E2%29%5Ex%2C+%22=%22%2C+%282%29%5E3%29, which becomes: matrix%281%2C3%2C+2%5E%282x%29%2C+%22=%22%2C+2%5E3%29

Now, 2x = 3, since the BASES are equal, and so, matrix%281%2C3%2C+x%2C+%22=%22%2C+3%2F2%29

We now see that: