Log4 7-log8 7+2log32 7 помогите, пожалуйста!

формула

$log_{a^k}b= \frac{1}{k}log_{a}b$

$log_{4}7=log_{2^2}7= \frac{1}{2}log_{2}7 \\ \\ log_{8}7=log_{2^3}7= \frac{1}{3}log_{2}7 \\ \\ log_{32}7=log_{2^5}7= \frac{1}{5}log_{2}7$

Формула
$n\cdot log_{a}b=log_{a}b^n$

$log_{4}7=log_{2^2}7= \frac{1}{2}log_{2}7=log_{2}7^{ \frac{1}{2} }=log_ {2} \sqrt{7} \\ \\ log_{8}7=log_{2^3}7= \frac{1}{3}log_{2}7 =log_{2} \sqrt[3]{7} \\ \\ 2log_{32}7=2log_{2^5}7= 2\cdot \frac{1}{5}log_{2}7=log_{2}7^{ \frac{2}{5} }$

Формулы
$log_{a}b-log_{a}c=log_{a} \frac{b}{c} \\ \\ log_{a} \frac{b}{c}+log_{a}d=log_{a} \frac{b}{c}\cdot d$

log₄ 7-log₈ 7+2log₃₂ 7=(1/2)log₂7-(1/3)log₂7+2·(1/5)log₂7=
=log₂√7-log₂∛7+log₂7²/⁵=log₂(√7/∛7)·7²/⁵=log₂7¹⁷/³⁰