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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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