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

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

Answer Link

Аноним Аноним · Answer 2 · 2018-05-18T23:26:28+02:00

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

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

[tex]\dfrac{1}{5}\log _9\left(\dfrac{a^6b}{81}\right)[/tex]

[tex]\log _9\left(\dfrac{a^6b}{81}\right)=\log _9\left(a^6b\right)-\log _9\left(81\right)[/tex]

[tex]\log _9\left(a^6b\right)-\log _9\left(81\right)[/tex]

[tex]\log _9\left(b\right)+\log _9\left(a^6\right)-2[/tex]

[tex]6\log _9\left(a\right)[/tex]

[tex]\dfrac{1}{5}\left(6\log _9\left(a\right)+\log _9\left(b\right)-2\right)[/tex]

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

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

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

[tex]\dfrac{1}{5}\cdot \:6\log _9\left(a\right)=\dfrac{6}{5}\log _9\left(a\right)[/tex]

[tex]\dfrac{1}{5}\log _9\left(b\right)+\dfrac{6}{5}\log _9\left(a\right)-\dfrac{2}{5}[/tex]

[tex]\dfrac{\log _9\left(b\right)}{5}+\log _9\left(a\right)\dfrac{6}{5}-\frac{2}{5}[/tex]

[tex]\log _9\left(a\right)\dfrac{6}{5}\::\quad \dfrac{6\log _9\left(a\right)}{5}[/tex]

[tex]\dfrac{\log _9\left(b\right)}{5}+\dfrac{6\log _9\left(a\right)}{5}-\dfrac{2}{5}[/tex]

[tex]\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}[/tex] [tex] =\ \textgreater \ \dfrac{-2+\log _9\left(b\right)+6\log _9\left(a\right)}{5}[/tex]

[tex]\text{Took a while, but here it is. Hope it helps!}[/tex]