math- precalculus

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how do i solve these?


9^(-x)=1/3



2^(2x)+ 2^(x) -12 = 0



(3/5)^x = 7 ^ (1-x)

  • math- precalculus -

    9^(-x)=1/3
    (3^2)^(-x) = 3^-1
    3^(-2x) = 3^-1
    so -2x = -1
    x = 1/2

    2^(2x)+ 2^(x) -12 = 0
    (2^x)^2 + 2^x - 12 = 0
    let y = 2^x

    y^2 + y - 12 = 0
    (y+4)(y-3) = 0
    y = -4 or y = 3

    2^x = -4, no solution , or
    2^x = 3 --- x = log3/log2

  • math- precalculus -

    last one:
    take logs of both sides

    (3/5)^x = 7 ^ (1-x)
    log(3/5)^x = log7 ^ (1-x)
    xlog(3/5) = (1-x)Log 7
    xlog .6 = log 7 - xlog 7
    xlog .6 + xlog 7 = log 7
    x(log .6 + log 7) = log 7
    x = log 7/(log.6 + log7)

  • math- precalculus -

    1/6

  • rwil ohsuycip -

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