Asked by kayla

how do i solve these logs?

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



3^(x^3)= 9^x
Answered by Dune
Wooah i don't get that ;]
do you understand mine?
Answered by Quidditch
Check your previous post for the solution to problem 2.
Answered by Reiny
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
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