Question

Помогите решить во вложении

Answer 1

A2.
$3^{log_2frac{1}{4}+log_35}=3^{log_22^{-2}}*3^{log_35}=3^{-2}*5=frac{5}{9}$


A3.
$log_6sqrt{a}=4\log_6a^frac{1}{2}=4\frac{1}{2}log_6a=4\log_6a=8\frac{1}{log_a6}=8\log_a6=frac{1}{8}=0.125$


A4.
$(12^{1-2log_{12}6}+7^{-2log_{49}6})^{-1}=(12*(12^{log_{12}6})^{-2}+(49^{log_{49}6})^{-1})^{-1}=\=(12*frac{1}{36}+frac{1}{6})^{-1}=(frac{2}{6}+frac{1}{6})^{-1}=2$

Answer 2

2)$3^{-2log(2)2} * 3^{log(3)5} = 3^{-2} *5=1/9*5=5/9$
3)log(6)√2=4⇒1/2log(6)a=4⇒log(6)a=8
log(a)6=1/log(6)a=1/8=0,125
4)$(12:12 ^{log(12)36} +49 ^{log(49)1/6} ) ^{-1} =(12:36+1/6) ^{-1} =(1/3+1/6) ^{-1} =$$(1/2) ^{-1} =2$