Respostas

2014-06-16T03:44:06-03:00
25^{x+2}=125^{x+5}

sabemos que 
25=5^2\\\\125=5^3

reescrevendo temos
(5^2)^{x+2}=(5^3)^{x+5}\\\\5^{2x+4} =5^{3x+15}\\\\\\\\2x+4=3x+15\\\\4=3x-2x+15\\\\4=x+15\\\\4-15=x\\\\-11=x

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5^{2x-1}=1

sabemos que todo numero elevado a 0 é igual a 1
então
2x-1=0\\\\2x=1\\\\x= \frac{1}{2}
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7^x=8^x
a unica maneira disso acontecer é quando x=0

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2*3^x = 162

tirando o mmc de 162
162 |2
81 | 3
27 | 3
9  | 3
3  | 3 
1
162 = 2*3*3*3*3\\\\162=2*3^4

2*3^x=2*3^4\\\\x=4

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3^{2x} +2*3^x -15 =0\\\\3^{2x}+2*3^x=15\\\\9^x+2*3^x=15

observando vemos que 9+6 = 15

pra vc ter 9+6 
então x =1

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A melhor resposta!
2014-06-17T00:45:50-03:00
Olá Juli,

aplique as propriedades da potenciação:

25^{x+2}=125^{x+5}\\
(5^2)^{x+2}=(5^3)^{x+5}\\
\not5^{2x+4}=\not5^{3x+15}\\
2x+4=3x+15\\
2x-3x=15-4\\
-x=11~~*~~(-1)\\
x=-11\\\\
\boxed{S=\{-11\}}

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5^{2x-1}=1\\
5 ^{2x-1}=5^0\\
\not5^{2x-1}=\not5^0\\
2x-1=0\\
2x=1\\\\
x= \dfrac{1}{2}\\\\\\
\boxed{S=\{ \dfrac{1}{2}\}}

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7^x=8^x\\\\
x=0

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2*3^x=162\\\\
3^x= \dfrac{162}{2}\\\\
3^x=81\\
3^x=3^4\\
\not3^x=\not3^4\\
x=4\\\\
\boxed{S=\{4\}}

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3^{2x}+2*3^x-15=0\\
(3^x)^{2}+2*3^x-15=0\\\\
fazendo~3^x=k\\\\
(k)^2+2*(k)-15=0\\
k^2+2k-15=0

\Delta=b^2-4ac\\
\Delta=2^2-4*1*(-15)\\
\Delta=4+60\\
\Delta=64

k= \dfrac{-b\pm \sqrt{\Delta} }{2a} \\\\\\
k= \dfrac{-2\pm \sqrt{64} }{2*1}= \dfrac{-2\pm8}{2}\begin{cases}k'= \dfrac{-2+8}{2}~\to~k'= \dfrac{6}{2}~\to~k'=3\\\\
k''= \dfrac{-2-8}{2}~\to~k''= \dfrac{-10}{2}~\to~k''=-5    \end{cases}\\\\\\
lembra,~3^x=k\\\\
3^x=3~~~~~~~~~~~3^x=-5~~(n\~ao~serve)\\
\not3^x=\not3^1\\
x=1\\\\
portanto:\\\\\\
\boxed{S=\{1\}}

espero ter ajudado e tenha ótimos estudos =))
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