Question

Resolva as seguintes equações: 

c) log2 ( log3 (x -1))= 2 

D)$log_{x+1} (x^{2} +7)=2$

e)$log_{2} 3+ log_{2}(x-1)= log _{2} 6$

f)$log_{3} 2+ log_{3}(x+1)=1$


g) 2log x  =log 2 +log x
 
 

h)$log_{2} ( x^{2} +2x-7)-log_{2}(x-1)=2$

Answer 1

Ilane,

dadas as equações logarítmicas, vamos usar as seguintes propriedades:

$logb+logc~\to~log(b*c)\\\\ k*logb~\to~logb^k\\\\ log_ab=k~\to~a^k=b\\\\ logb-logc~\to~log \dfrac{b}{c}$

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Condições para que log exista:

$\begin{cases}x>0~~e~~x \neq 1~\to~para~a~base\\ x>0~\to~para~o~logaritmando\end{cases}$

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$log_2[(log_3(x-1)]=2\\\\ log_3(x-1)=2^2\\ log_3(x-1)=4\\ x-1=3^4\\ x-1=81\\ x=81+1\\ x=82\\\\ \boxed{S=\{82\}}$

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$log_{x+1}( x^{2} +7)=2\\\\ (x+1)^2= x^{2} +7\\ \not{x^{2}} +2x+1= \not{x^{2}} +7\\ 2x+1=7\\ 2x=7-1\\ 2x=6\\\\ x= \dfrac{6}{2}\\\\ x=3\\\\ \boxed{S=\{3\}}$

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$log_23+log_2(x-1)=log_26\\\\ bases~iguais,~elimine-as,~e~aplique~a~1a~propriedade~de~log\\\\ 3*(x-1)=6\\ 3x-3=6\\ 3x=6+3\\ 3x=9\\\\ x= \dfrac{9}{3}\\\\ x=3\\\\ \boxed{S=\{3\}}$

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$log_32+log3(x+1)=1\\ log_3[2*(x+1)]=log_33\\ 2*(x+1)=3\\ 2x+2=3\\ 2x=3-2\\ 2x=1\\\\ x= \dfrac{1}{2}\\\\ \boxed{S=\left\{ \dfrac{1}{2}\right\}}$

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$2logx=log2+logx\\ logx^2=log(2*x)\\\\ x^{2} =2x\\ x^{2} -2x=0\\ x(x-2)=0\\\\ x'=0~~x''=2\\\\ como~x=0~n\~ao~atende~\`a~condic\~ao~de~exist\^encia,~temos~que:\\\\ \boxed{S=\{2\}}$

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$log_2( x^{2} +2x-7)-log_2(x-1)=2\\\\ log_2 \left(\dfrac{ x^{2} +2x-7}{x-1}\right)=2\\\\ como~2=log_24,~temos~que:\\\\ log_2\left( \dfrac{ x^{2} +2x-7}{x-1}\right)=log_24\\\\\\ \dfrac{ x^{2} +2x-7}{x-1}=4\\\\\\ x^{2} +2x-7=4(x-1)\\ x^{2} +2x-7=4x-4\\ x^{2} -2x-3=0$

$\Delta=b^2-4ac\\ \Delta=(-2)^2-4*1*(-3)\\ \Delta=4+12\\ \Delta=16$

$x= \dfrac{-b\pm \sqrt{\Delta} }{2a}\\\\\\ x= \dfrac{-(-2)\pm \sqrt{16} }{2*1}= \dfrac{2\pm4}{2}\begin{cases}x'= \dfrac{2-4}{2}\to~x'= \dfrac{-2}{~~2}\to~x'=-1\\\\ x''= \dfrac{2+4}{2}\to~x''= \dfrac{6}{2}\to~x''=3\end{cases}$

Como x= -1 não atende à condição de existência, a solução da equação é:

$\boxed{S=\{3\}}$

Espero ter ajudado e tenha ótimos estudos =))