Question

Resolva a equação: $2 ^{x-1} + 2 ^{x+3} +2 ^{x-2} + 2 ^{x} = 2496$
a) x=8
b)x=6
c)x=7
d)x=9
e)x=-7

Preciso urgente

Answer 1

$\frac{2^x}{2} +2^x.2^3 + \frac{2^x}{2^2} + 2^x = 2496$

Agora faz: 2^x = y

$\frac{y}{2} +y.2^3 + \frac{y}{2^2} + y = 2496$

$\frac{y}{2} +8y + \frac{y}{4} + y = 2496$

multiplica tudo por 4:

$2y +32y + y + 4y = 9984$

$39y = 9984$

$y = 256$

então:


$2^x = 256$

$2^x = 2^8$

$x = 8$

Answer 2

$2^{x-1} = 2^{x}. 2^{-1} = \frac{1}{2} . 2^{x}$

$2^{x+3} = 2^{x} . 2^{3} =8. 2^{x}$

$2^{x-2} = 2^{x} . 2^{-2} = \frac{1}{4} . 2^{x}$

$2^{x} =2^{x}$

$\frac{1}{2} 2^{x} +8. 2^{x} + \frac{1}{4} . 2^{x} + 2^{x} =2496$

$2^{x} (\frac{1}{2} +8+ \frac{1}{4}+ 1) =2496$

$2^{x} (\frac{2+32+1+4}{4} ) =2496$

$2^{x} (\frac{39}{4} ) =2496$

$2^{x} = \frac{4}{39} .2496$

$2^{x} =256$

$2^{x} = 2^{8}$

$x=8$