Question 375676
Given 
{{{log(a, (2)) =x}}}
and
{{{log(16,(a^log(a, (x))))=100}}}<br>
Let's start by looking at: {{{a^log(a, (x))}}}. Once you understand logarithms well you will see instantly how this simplifies. The exponent, {{{log(a, (x))}}}, 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^log(a, (x))}}} is x! The second equation is now:
{{{log(16,(x))=100}}}
Into this equation we can substitute for x using the first equation:
{{{log(16,(log(a, (2))))=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(z, (p)) = q}}} is equivalent to {{{p = z^q}}}. Using this on the equation above we get:
{{{log(a, (2))= 16^100}}}
Rewriting this equation in exponential form we get:
{{{2 = a^(16^100)}}}
And last we find the {{{16^100}}}th root of each side:
{{{root((16^100), 2) = 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 {{{16^100}}}th root of 2.)<br>
The {{{16^100}}}th 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.