Question

49〖^(x+1)〗=∛7 alguem sabe?

Answer 1

$49^{x+1}= \sqrt[3]{7} = (7^{2})^{x+1} = 7^{ \frac{1}{3} }=7^{2x+2}=7^{ \frac{1}{3} } \\ \\ 2x + 2 = \frac{1}{3} = 6x + 6 = 1 \\ \\ 6x = 1 - 6 \\ \\ 6x = -5 \\ \\ x = - \frac{5}{6}$

Espero ter ajudado.