Question

Помогите решить!
ответ: x1=㏒₂(40); x2=3

Answer 1

ОДЗ:
х+3≠0
х≠-3


$625=5^4 \\ \\ 1600=4^3*5^2$

решение:
$4^x*625^ \frac{x}{x+3} =1600 \\ \\ 4^x*(5^4)^ \frac{x}{x+3} =4^3*5^2 \\ \\ 4^x*5^ \frac{4x}{x+3} =4^3*5^2$

$\frac{5^ \frac{4x}{x+3} }{5^2} = \frac{4^3}{4^x} \\ \\ 5^{ \frac{4x}{x+3} -2}=4^{3-x} \\ \\ 4^{3-x}=5^{ \frac{2x-6}{x+3} } \\ \\ log_4(4^{3-x})=log_4(5^{ \frac{2x-6}{x+3}})\\ \\ 3-x= \frac{2x-6}{x+3}}log_{2^2}5 \\ \\ 3-x= \frac{2(x-3)}{2(x+3)}}log_25 \\ \\ 3-x= \frac{x-3}{x+3} log_25 \ \ |*(3+x) \\ \\ (3+x)(3-x)=(x-3)log_25 \\ \\ 9-x^2=xlog_25-3log_25$

$9-x^2=xlog_25-3log_25 \\ \\ x^2+xlog_25-(3log_25+9)=0 \\ \\ D=log^2_25+4(3log_25+9)=log^2_25+12log_25+36 \\ \\ \sqrt{D} = \sqrt{ log^2_25+12log_25+36}= \sqrt{ (log_25+6)^2}=log_25+6 \\ \\ x_{1,2}= \frac{ -b^+_-\sqrt{D}}{2a} \\ \\ x_1= \frac{-log_25+(log_25+6)}{2} = \frac{-log_25+log_25+6}{2} = \frac{6}{2}=3 \\ \\ x_2= \frac{-log_25-(log_25+6)}{2} = \frac{-log_25-log_25-6}{2} = \frac{-2log_25-6}{2} =-log_25-3= \\ \\ =-log_25-3log_22=-log_25-log_22^3=-log_25-log_28=\\ \\ =-(log_25+log_28)=-log_240$


$OTBET: \ 3; -log_240$