Ребятки , дорогие,очень надо❤❤❤

Question 1

Question 2

$log_{0.05}(40 \sqrt{5} )=log_{ \frac{1}{20} }( \sqrt{40^2*5} )=-log_{20} \sqrt{8000}=- \frac{1}{2} log_{20}20^3=$ $=-0.5*3 =-1.5$
1)
$log_b( \sqrt[5]{a} * \sqrt[4]{b} ),$ если $log_b \sqrt[9]{a} =4$

$log_b {a^{ \frac{1}{9}} } =4$
$\frac{1}{9}} log_b {a =4$
$log_b {a =36$
$log_b( \sqrt[5]{a} * \sqrt[4]{b} )=log_b\sqrt[5]{a} +log_b \sqrt[4]{b} =log_b{a^{ \frac{1}{5} }} +log_b {b^{ \frac{1}{4} }}= \frac{1}{5} log_ba+ \frac{1}{4}$

$\frac{1}{5} log_ba+ \frac{1}{4} =\frac{1}{5}*36+ \frac{1}{4} =7.2+0.25=7.45$
2)
$log_{32}log_{324}(3 \sqrt[4]{4})= log_{32}log_{324}( \sqrt[4]{3^4*4})= log_{32}log_{324}( \sqrt[4]{324})=$ $= log_{32}log_{324}( 324^{ \frac{1}{4} })= log_{32} \frac{1}{4}= log_{2^5} 2^{-2}=-2* \frac{1}{5} =-0.4$
3)
$log_{49} (7 \sqrt[4]{343}) =log_{7^2} (7 \sqrt[4]{7^3}) = \frac{1}{2} log_7(7*7^{ \frac{3}{4} })=0.5log_77^{1.75}=$ $=0.5*1.75=0.875$
4)
$log_{49}8*log_25*log_{25}7=log_{49}2^3*log_25*log_{25}7=$ $=3log_{49}2*log_25*log_{25}7=3log_{49}5*log_{5^2}7=3* \frac{1}{2} log_{49}5*log_{5}7=$ $=1.5log_{49}7=1.5log_{7^2}7=1.5*0.5=0.75$
5)
$log^2_296*log_{12}2- \frac{(log_212-3)^2}{log_212} =log^2_2(12*8)* \frac{1}{log_212} - \frac{(log_212-3)^2}{log_212}$ $=(log_212+log_28)^2* \frac{1}{log_212} - \frac{(log_212-3)^2}{log_212}$
Замена: $log_212=a$
$=(a+3)^2* \frac{1}{a} - \frac{(a-3)^2}{a}= \frac{(a+3)^2}{a} -\frac{(a-3)^2}{a}=\frac{(a+3)^2-(a-3)^2}{a}=$ $=\frac{(a+3+a-3)(a+3-a+3)}{a}= \frac{2a*6}{a} =12$