# precalculus

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how do i solvethis equation (btw we are studyinglogarithms)

3^(x^3)=9^x

2^x=10

(4^x)-(2^x)=0

• precalculus -

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

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