Question

pergunta de logaritimo urgente!

Answer 1

Vamos là.

log32(2^(x - 3)) = x

log(2^(x - 3)) = x*log(32)

(x - 3)*log(2) = 5xlog(2)

x - 3 = 5x

4x = -3

x = -3/4

Answer 2

Resposta:

$x=-\frac{3}{4}$

Explicação passo a passo:

$log_3_2~2^x^-^3=x$

Dodo, logₐ b = c, devemos ter: b > 0 e 0 < a ≠ 1

b = a^c

$C.E: 2^x^-^3>0\implies x \in R\\\\2^x^-^3=32^x\\\\2^x^-^3=(2^5)^x\\\\2^x^-^3=2^5^x\\\\5x=x-3\\\\5x-x=-3\\\\4x=-3\\\\x=-\frac{3}{4}$