Question

QF Q7.) Use properties of logarithms to expand the logarithmic expression as much as possible. Evaluate logarithmic expressions without using a calculator if possible.

Answer 1

Attached is the solution.
The following log properties are used:
$log(x^n) = nlog(x)$
$log(\frac{a}{b}) = log(a) - log(b) \\ \\ log(ab) = log(a) + log(b) \\ \\ log_9 (81) = 2 \rightarrow 9^2 = 81$

Answer 2

$\log _9\left(\sqrt[5]{\dfrac{a^6b}{81}}\right)=\dfrac{-2+\log _9\left(b\right)+6\log _9\left(a\right)}{5}$ 

$\log _9\left(\sqrt[5]{\dfrac{a^6b}{81}}\right) =\ \textgreater \ \log _9\left(\left(\dfrac{a^6b}{81}\right)^{\dfrac{1}{5}}\right)$ 

$\dfrac{1}{5}\log _9\left(\dfrac{a^6b}{81}\right)$ 

$\log _9\left(\dfrac{a^6b}{81}\right)=\log _9\left(a^6b\right)-\log _9\left(81\right)$ 

$\log _9\left(a^6b\right)-\log _9\left(81\right)$ 

$\log _9\left(b\right)+\log _9\left(a^6\right)-2$ 

$6\log _9\left(a\right)$ 

$\dfrac{1}{5}\left(6\log _9\left(a\right)+\log _9\left(b\right)-2\right)$ 

$\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\left(a\right)+\dfrac{1}{5}\left(-2\right)$ 

$\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\dleft(a\right)-\frac{1}{5}\cdot \:2)$ 

$\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\left(a\right)-\dfrac{1}{5}\cdot \:2$ 

$\dfrac{1}{5}\cdot \:6\log _9\left(a\right)=\dfrac{6}{5}\log _9\left(a\right)$ 

$\dfrac{1}{5}\log _9\left(b\right)+\dfrac{6}{5}\log _9\left(a\right)-\dfrac{2}{5}$ 

$\dfrac{\log _9\left(b\right)}{5}+\log _9\left(a\right)\dfrac{6}{5}-\frac{2}{5}$ 

$\log _9\left(a\right)\dfrac{6}{5}\::\quad \dfrac{6\log _9\left(a\right)}{5}$ 

$\dfrac{\log _9\left(b\right)}{5}+\dfrac{6\log _9\left(a\right)}{5}-\dfrac{2}{5}$ 

$\dfrac{\log _9\left(b\right)}{5}+\dfrac{6\log _9\left(a\right)}{5}:\quad \dfrac{\log _9\left(b\right)+6\log _9\left(a\right)}{5}$ $=\ \textgreater \ \dfrac{-2+\log _9\left(b\right)+6\log _9\left(a\right)}{5}$ 

$\text{Took a while, but here it is. Hope it helps!}$