Question

1. log(2)56-log(2)7+16 log(2)3

2. log(7)(3x-5)-log(7)(9-2x)=1

3. 4-lg^2x=3tgx 4. log 13(2-3x)<-2

ренить систему

log(2)(x-y)+2log(4)(x-y)=3

3(2)+log(3)(2x-y)=45

Answer 1

$log_256-log_27+16 log_23= log_2(56/7)+log_2(3^1^6)= \ =log_28+log_2(3^1^6)=3+16log_23 \ \ 2.\ log_7(3x-5)-log_7(9-2x)=1 \log_7frac{3x-5}{9-2x}=log_77 \ frac{3x-5}{9-2x}=7 \ frac{3x-5-7(9-2x)}{9-2x}=0 \ frac{3x-5-63+14x}{9-2x}=0 \ 17x=68 => x=4\ \ 3.\ 4-lg^2x=3lgx \ t=lgx\ 4-t^2-3t = 0 \ t^2+3t-4=0 \ (t+4)(t-1)=0\lgx_1=1 <=> x=10 \ lgx_2=-4 <=> x=10^{-4} \ \ 4.\ log _{1/3}(2-3x)<-2 \ -log _{3}(2-3x)<-log_39 \ log _{3}(2-3x)>log_39 \ (2-3x)>9 \ 3x<11 => x in (- infty; 11/3)$

Answer 2

первое задание возможно выглядит так log(2)56-log(2)7+16^log(2)3=Log(2)(56/7)+2^(4log(2)3)=3+3^4=84