how do i solve these logs?

log1/2 (3x+1)^1/3= -2

3^(x^3)= 9^x

3 answers

Wooah i don't get that ;]
do you understand mine?
Check your previous post for the solution to problem 2.
first problem

convert your equation to exponential form
(1/2)^-2 = (3x+1)^1/3
2^2 = (3x+1)^1/3
4 = (3x+1)^1/3 , cube both sides
64 = 3x+1
63 = 3x
x = 21