SOLUTION: Find each value of x. {{{2^log2^5=x}}}.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find each value of x. {{{2^log2^5=x}}}.      Log On


   

Question 278548: Find each value of x.
2%5Elog2%5E5=x.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%5Elog2%5E5+=+x
If this is what the problem really looks like then, since 2%5E5+=+32, this simplifies to:
2%5Elog%28%2832%29%29=x
Now we use our calculators to find log(32) and then use the calculators again to raise to to that power.

However, I suspect the the problem is actually:
2%5Elog%282%2C+%285%29%29+=+x
This equation is extremely easy if you understand what logairthms are. Logarithms are exponents. In general, log%28a%2C+%28b%29%29 represents the exponent you put on a which results in b. This specific logarithm, log%282%2C+%285%29%29, represents the exponent you put on 2 to get 5. And where do we see log%282%2C+%285%29%29? We see it as an exponent on 2! So 2%5Elog%282%2C+%285%29%29, by the definition of logarithms, is 5!