[tex] log_{8} \frac{7+5x}{5x-7} = log_{4} \frac{3+2x}{3-2x}[/tex]решить уравнение. Created by атив1. algebra-ru

tex]решить уравнение...

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$log_{8} \frac{7+5x}{5x-7} = log_{4} \frac{ 3+2x}{3-2x}\\\\
 log_{8}(7+5x)-log_{8}(5x-7)=log_{4}(3+2x)-log_{4}(3-2x) \\\\
 \frac{log_{2}(7+5x)}{3} - \frac{log_{2}(5x-7)}{3}=\frac{log_{2}(3+2x)}{2}-\frac{ log_{2}(3-2x)}{2}\\\\
2log_{2}(7+5x)-2log_{2}(5x-7)=3log_{2}(2x+3) -3log_{2}(3-2x)\\\\
 log_{2}\frac{(7+5x)^2}{ (5x-7)^2 }=log_{2}\frac{(2x+3)^3}{(3-2x)^3}\\\\
(\frac{7+5x}{5x-7})^2 = \frac{(2x+3)^3}{(3-2x)^3}\\\\
(x^2-2)*4x(100x^2-189)=0\\\\
 x=+-\sqrt{2}\\
 x=0\\
 x=+-\frac{ \sqrt{189}}{10}$
Подходит
$+-\sqrt{2}$

2 votes Thanks 1

Gerren Log8((7+5x)/(5x-7))=log4((3+2x)/(3-2x)) одз (7+5x)/(5x-7)>0
1/3log2((7+5x)/(5x-7))=1/2log2((3+2x)/(3-2x)) + -7/5 - 7/5 +
log2((7+5x)/(5x-7))^1/3)=log2((3+2x)/(3-2x))^1/2) (3+2x)/(3-2x)>0
(7+5x)/(5x-7))^1/3=(3+2x)/(3-2x))^1/2 + -3/2 - 2/3 +
(7+5x)/(5x-7))^2=(3+2x)/(3-2x))^3
(7+5x)/(5x-7))^2-(3+2x)/(3-2x))^3=0
(4х(х^2-2)(100х^2-189))/((2х-3)^3(5х-7)^2)=0
4x=0 x=0 посторонний корень
x^2-2=0
x^2=2
x=-+√2 (≈1,4142) подходят
100х^2-189=0
x^2=189/100
x^2=1,89
x≈+-1,374 посторонние корни
ответ
x=-√2 x=√2

1 votes Thanks 1

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