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Pre-Calculus

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I'm having problems understanding how to do these problems.

25. 32^(2x-3)=2

27. 1/4= 2^3x

29. 9^x=27

33. 81^(x-1)= 27^2x

  • Pre-Calculus - ,

    recall that log(x^a) = a * log x

    25)
    32^(2x-3) = 2
    take log of both sides to get
    (2x-3) log 32 = log 2
    2x-3 = log 2 / log 32
    Now recall that loga/logb = log_b(a)
    so, log2/log32 = log_32(2) = 1/5
    since 2^5 = 32, 2 = 32^(1/5), so log_32(2) = 1/5

    2x-3 = 1/5
    2x = 16/5
    x = 8/5

    Or, we could have started out by noting that 2 = 32^(1/5), so equating powers of 32,
    2x-3 = 1/5
    ...

    27)
    similarly, since 1/4 = 2^-2,
    -2 = 3x
    x = -2/3

    29)
    Since 3= 9^1/2, 27 = 3^3 = 9^3/2, so
    x = 3/2

    33)
    27 = 3^3
    81 = 3^4, so
    27 = 81^(3/4)
    equating powers,
    x-1 = 2x(3/4)
    x-1 = 3x/2
    x/2 = -1
    x = -2
    check:
    81^-3 = 3^-12
    27^-6 = 3^-12

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