precalculus
posted by kayla .
how do i solvethis equation (btw we are studyinglogarithms)
3^(x^3)=9^x
2^x=10
(4^x)(2^x)=0

for the first you don't even need logs
3^(x^3)=9^x
3^(x^3)=(3^2)^x
3^(x^3) = 3^(2x)
so x^3 = 2x
x^3  2x = 0
x(x^2  2) = 0
x = 0 or x = ±√2
for second, take logs of both sides
log(2^x) = log 10
xlog2 = 1
x = 1/log2
for the last
4^x = 2^x
2^(2x) = 2^x
2x = x
x = 0