SOLUTION: Given log(base"a")2=x, determine log(base"16)(a^log(base"a")x)=100. Solve for a. I have simplified the question to (4x)(a^log(base"a")x)=100. I am not sure what I can now do to

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Given log(base"a")2=x, determine log(base"16)(a^log(base"a")x)=100. Solve for a. I have simplified the question to (4x)(a^log(base"a")x)=100. I am not sure what I can now do to       Log On


   



Question 375676: Given log(base"a")2=x, determine log(base"16)(a^log(base"a")x)=100. Solve for a.
I have simplified the question to (4x)(a^log(base"a")x)=100. I am not sure what I can now do to solve for "a".

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Given
log%28a%2C+%282%29%29+=x
and
log%2816%2C%28a%5Elog%28a%2C+%28x%29%29%29%29=100

Let's start by looking at: a%5Elog%28a%2C+%28x%29%29. Once you understand logarithms well you will see instantly how this simplifies. The exponent, log%28a%2C+%28x%29%29, is a logarithm. Logarithms are exponents. This logarithm represents "the exponent for "a" that results in "x". And where do we find this exponent? Answer: As the exponent for "a"! So by definition, a%5Elog%28a%2C+%28x%29%29 is x! The second equation is now:
log%2816%2C%28x%29%29=100
Into this equation we can substitute for x using the first equation:
log%2816%2C%28log%28a%2C+%282%29%29%29%29=100
We now have an equation with just a in it. We can now solve for a, First we rewrite the equation in exponential form. In general log%28z%2C+%28p%29%29+=+q is equivalent to p+=+z%5Eq. Using this on the equation above we get:
log%28a%2C+%282%29%29=+16%5E100
Rewriting this equation in exponential form we get:
2+=+a%5E%2816%5E100%29
And last we find the 16%5E100th root of each side:
root%28%2816%5E100%29%2C+2%29+=+a
(Since "a" was the base of a logarithm it had to be positive. So we do not need to be concerned with the negative 16%5E100th root of 2.)

The 16%5E100th root of 2 is a very strange answer. Perhaps you made a mistake in posting this problem. I have checked my work and I do not see any mistakes.