Question

ME AJUDEM FAÇAM AS QUE CONSEGUIREM 

Answer 1

Olá Lara,

use as propriedades:

$loga+logb~\to~log(a*b)\\\\ loga-logb~\to~log\dfrac{a}{b}\\\\ logb^k~\to~k*logb\\\\ log_bn=k~\to~n=b^k\\\\ log_kk=1$

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Como as incógnitas estão no logaritmando, vamos impor a sua condição:

$x>0$

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$log_4(x+ x^{2} )= \dfrac{1}{2}\\\\ x+ x^{2} =4^{ \tfrac{1}{2} }\\ x^{2} +x=2\\ x^{2} +x-2=0\\ (x-1)(x+2)=0\\\\ x'=1~~~~~~x''=-2\\\\ \boxed{S=\{1,-2\}}$

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$log_2(12-2^x)=2x\\ 2^{2x}=12-2^x\\ (2^x)^2=12-2^x\\\\ 2^x=b\\\\ b^2=12-b\\ b^2+b-12=0\\ (b-3)(b+4)=0\\\\ b'=3~~~~~~e~~~~~~b''=-4~(n\~ao~serve)\\\\ 2^x=b\\ 2^x=3\\ log2^x=log3\\ x*log2=log3\\ x*0,3010=0,477\\\\ x= \dfrac{0,477}{0,301}\\\\ \boxed{x\approx1,58}$

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$log_2(4-x)=log_2(x+1)+1\\ log_2(4-x)=log_2(x+1)+log_22\\ 4-x=(x+1)*2\\ 4-x=2x+2\\ -x-2x=2-4\\ -3x=-2\\\\ x= \dfrac{-2}{-3}\\\\ \boxed{x= \dfrac{2}{3}}$

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$log_2(4-x)-log_2(x+1)=1\\\\ log_2( \dfrac{4-x}{x+1})=1\\\\ \dfrac{4-x}{x+1}=2^1\\\\ \dfrac{4-x}{x+1}=2\\\\ 4-x=2(x+1)\\ 4-x=2x+2\\ -x-2x=2-4\\ -3x=-2\\\\ x= \dfrac{-2}{-3}\\\\ \boxed{x= \dfrac{2}{3}}$

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$-1=log_5( \dfrac{2x}{x+1})\\\\ log_5( \dfrac{2x}{x+1})=-1\\\\ \dfrac{2x}{x+1}=5^{-1}\\\\ \dfrac{2x}{x+1} = \dfrac{1}{5}\\\\ 5*2x=x+1\\ 10x=x+1\\ 10x-x=1\\ 9x=1\\\\ \boxed{x= \dfrac{1}{9}}$

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$logx=log(3x+5)\\ x=3x+5\\ x-3x=5\\ -2x=5\\\\ x=- \dfrac{5}{2}~\to~n\~ao~atende~a~condic\~ao~de~exist\^encia\\\\ \boxed{S=\{\varnothing\}}$

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$log2+log(x+1)-logx=1\\ log2*(x+1)-logx=1\\\\ log( \dfrac{2x+2}{x})=1\\\\ log_{10}( \dfrac{2x+2}{x})=1\\\\ \dfrac{2x+2}{x}=10^1\\\\ \dfrac{2x+2}{x}=10\\\\ 2x+2=10*x\\ 2x+2=10x\\ 10x-2x=2\\ 8x=2\\\\ x= \dfrac{2}{8}\\\\ \boxed{x= \dfrac{1}{4}}$


Espero ter ajudado e tenha ótimos estudos =))