Question

(log2)^2(4-x)+log1/2(8/(4-x)) = 2^log4(9) ни как не могу решить,

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 $y = log_2(4-x)$

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

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Использованы формулы

$log_{\frac{1}{a}} b=-log_ab=log_ab^{-1}=log_a\dfrac{1}{b}\\ \\ log_{a^n}b^n=log_ab;~~a,b0,~~a\neq 1\\ \\ log_ab=c~~~\Rightarrow~~~a^c=b$