Question 1006514
Let ln (a)=2, ln (b)=3, ln (c)=5. Solve:
f (x)=(ln (c^2))(ln (a^4/b^2))^-1
<pre>ln a = 2
ln b = 3
ln c = 5
{{{f(x) = (ln (c^2))(ln (a^4/b^2))^(- 1)}}}
{{{ln (c^2) = 2 ln (c)}}} ------- Applying {{{ln (p^q) = q ln (p)}}} 
{{{(ln (a^4/b^2))^(- 1) = 1/(ln (a^4/b^2))}}}
We therefore have: 
{{{f(x) = 2 ln (c) * (1/(ln (a^4/b^2)))}}}
{{{f(x) = 2 ln (c) * (1/(ln (a^4) - ln (b^2)))}}} --- Applying {{{ln (p/q)}}} = ln p – ln q
{{{f(x) = 2 ln (c) * (1/(4 ln (a) - 2 ln (b)))}}} ---- Applying {{{ln (p^q)}}} = q ln p
{{{f(x) = 2(5) * (1/(4(2) - 2(3)))}}} ----------- Substituting 5 for ln c, 2 for ln a, and 3 for ln b
{{{f(x) = 10 * (1/(8 - 6))}}}
{{{f(x) = 10 * (1/2)}}}, or {{{highlight_green(5)}}}