Question

(Log2)^2(4-x)+log1/2(8/(4-x)) = 2^log4(9) ни как не могу решить, помогите

Answer 1

$displaystyle log_2^2(4-x)+log_{frac{1}{2}}(frac{8}{4-x}) = 2^{log_49}$

ОДЗ : 4-x > 0; x < 4

$displaystyle log_2^2(4-x)+log_{2}(frac{8}{4-x})^{-1} = 2^{log_{2^2}3^2}\ \ displaystyle log_2^2(4-x)+log_{2}(frac{4-x}{8}) = 2^{log_23}\ \ displaystyle log_2^2(4-x)+log_{2}(4-x) - log_28= 3\ \ displaystyle log_2^2(4-x)+log_{2}(4-x) - 6=0$

Замена $y = log_2(4-x)$

$y^2+y-6=0; ~~~(y+3)(y-2)=0\ \1)~~y+3=0;~~y=-3\~~~~log_2(4-x)=-3;\~~~~ 2^{-3}=4-x\ ~~~~x=4-dfrac{1}{8}=3dfrac{7}{8};~~~~boxed{boldsymbol{x_1=3,875}}\ \ 2)~~y-2=0; ~~y=2\ ~~~~log_2(4-x)=2;\~~~~2^2=4-x \ ~~~~x = 4-4=0~~~~boxed{boldsymbol{x_2=0}}$

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Использованы формулы

$log_{frac{1}{a}} b=-log_ab=log_ab^{-1}=log_adfrac{1}{b}\ \ log_{a^n}b^n=log_ab;~~a,b>0,~~aneq 1\ \ log_ab=c~~~Rightarrow~~~a^c=b$