Question

resolva as seguintes equações exponenciais apresentando a solução na variedade x :
a) 25 elevado a x+2 =125 elevado a x+5

b)5 elevado a 2x-1= 1

c)7×=8×

d)2.3×=162

e)3 elevado a 2×+2.3×-15=0

Answer 1

$25^{x+2}=125^{x+5}$

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

Answer 2

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 =))