Question

Determine x, com x € R, em cada equação:

a) 128^x+2 = 512

b) 2^x+1 = 128

c) 3^x-1 + 3^x+2 - 3^x = 25

d) 4^x = 7.2^x + 8​

Answer 1

Explicação passo-a-passo:

a)

$\sf 128^{x+2}=512$

$\sf (2^7)^{x+2}=2^9$

$\sf 2^{7x+14}=2^9$

Igualando os expoentes:

$\sf 7x+14=9$

$\sf 7x=9-14$

$\sf 7x=-5$

$\sf \red{x=\dfrac{-5}{7}}$

b)

$\sf 2^{x+1}=128$

$\sf 2^{x+1}=2^7$

Igualando os expoentes:

$\sf x+1=7$

$\sf x=7-1$

$\sf \red{x=6}$

c)

$\sf 3^{x-1}+3^{x+2}-3^x=25$

$\sf 3^{x-1}\cdot(1+3^3-3)=25$

$\sf 3^{x-1}\cdot(1+27-3)=25$

$\sf 3^{x-1}\cdot25=25$

$\sf 3^{x-1}=\dfrac{25}{25}$

$\sf 3^{x-1}=1$

$\sf 3^{x-1}=3^0$

Igualando os expoentes:

$\sf x-1=0$

$\sf \red{x=1}$

d)

$\sf 4^x=7\cdot 2^{x}+8$

$\sf (2^2)^x=7\cdot 2^{x}+8$

$\sf (2^x)^2=7\cdot 2^{x}+8$

$\sf (2^x)^2-7\cdot 2^{x}-8=0$

Seja $\sf 2^x=y$

$\sf y^2-7y-8=0$

$\sf \Delta=(-7)^2-4\cdot1\cdot(-8)$

$\sf \Delta=49+32$

$\sf \Delta=81$

$\sf y=\dfrac{-(-7)\pm\sqrt{81}}{2\cdot1}=\dfrac{7\pm9}{2}$

$\sf y'=\dfrac{7+9}{2}~\Rightarrow~y'=\dfrac{16}{2}~\Rightarrow~\green{y'=8}$

$\sf y"=\dfrac{7-9}{2}~\Rightarrow~y"=\dfrac{-2}{2}~\Rightarrow~y"=-1$ (não serve)

Assim:

$\sf 2^x=8$

$\sf 2^{x}=2^3$

Igualando os expoentes:

$\sf \red{x=3}$