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h) i) Solve for x in 27^x = 93x−2

Mathematics

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4 steps

Part h(i)· 4 steps

27x=93x−227^x = 9^{3x-2}

Solve for xx in

  1. Express both bases as powers of the same number, 3.

    (33)x=(32)3x−2(3^3)^x = (3^2)^{3x-2}
  2. Apply the power rule (am)n=amn(a^m)^n = a^{mn}.

    33x=32(3x−2)3^{3x} = 3^{2(3x-2)} 33x=36x−43^{3x} = 3^{6x-4}
  3. Equate the exponents since the bases are the same.

    3x=6x−43x = 6x-4
  4. Solve for xx.

    3x−6x=−43x - 6x = -4 −3x=−4-3x = -4 x=−4−3x = \frac{-4}{-3} x=43x = \frac{4}{3}

    The solution for h) i) is $\boxed{x =

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