Question 959136
First we will find x and then x^(-1/2). To find x we will repeatedly rewrite the equation in exponential form (to "peel away" the logs).<br>
In general {{{log(a, (p)) = n}}} is equivalent to {{{a^n = p}}}. Using this pattern on
{{{log(7, (log(3, (log(2, (x))))))=0}}}
{{{7^0 = log(3, (log(2, (x))))))}}}
which simplifies to:
{{{1 = log(3, (log(2, (x))))))}}}
Repeating...
{{{3^1 = log(2, (x))))}}}
{{{3 = log(2, (x))))}}}
Again...
{{{2^3 = x}}}
{{{8 = x}}}<br>
Now for {{{x^(-1/2)}}}:
{{{8^(-1/2)}}}
{{{1/8^(1/2)}}}
{{{1/sqrt(8)}}}
{{{(1/sqrt(8))(sqrt(2)/sqrt(2))}}}
{{{sqrt(2)/sqrt(16)}}}
{{{sqrt(2)/4}}}