Помогите решить 80 баллов а)7*3^x+1-3^x+4=5^x+2-5^x+3 б)2^x-2^x+2+2^x-1=(3^x+1-3^x+2+3^x)*2/9. Created by Andylov. algebra-ru

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Andylov @Andylov

August 2022 1 25 Report

Помогите решить 80 баллов
а)7*3^x+1-3^x+4=5^x+2-5^x+3
б)2^x-2^x+2+2^x-1=(3^x+1-3^x+2+3^x)*2/9

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90misha90

Verified answer

$7*3^{x+1}-3^{x+4}=5^{x+2}-5^{x+3}$
$7*3*3^{x}-3^4*3^{x}=25*5^{x}-125*5^{x}$
$-60*3^{x}=-100*5^{x}$
$-60*3^{x}=-100*5^{x}$
$3*3^{x}=5*5^{x}$
$3^{x+1}=5^{x+1}$
$( \frac{3}{5}) ^{x+1}=1$
$( \frac{3}{5}) ^{x+1}= (\frac{3}{5} )^0$
$x+1=0$

$x=-1$
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$2^x-2^{x+2}+2^{x-1}=(3^{x+1}-3^{x+2}+3^x)* \frac{2}{9}$
$2^x-4*2^{x}+ \frac{1}{2}* 2^{x}=(3*3^{x}-9*3^{x}+3^x)* \frac{2}{9} |*2$
$2*2^x-8*2^{x}+ 2^{x}=(3*3^{x}-9*3^{x}+3^x)* \frac{4}{9}$
$-5*2^{x}=(-5*3^x)* \frac{4}{9}$
$2^{x}=3^x *(\frac{2}{3})^2$
$(\frac{2}{3})^x =(\frac{2}{3})^2$

$x=2$

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