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Express all numbers in the equation as powers of the same base. The base 2 is suitable.

Chemistry

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  1. Express all numbers in the equation as powers of the same base.

    The base 2 is suitable.

    2=212 = 2^1 8=238 = 2^3 128=27128 = 2^7

  2. Substitute these into the given equation.

    2(x−3)×(23)(x+2)=272^{(x-3)} \times (2^3)^{(x+2)} = 2^7

  3. Apply the exponent rule (am)n=amn(a^m)^n = a^{mn} to the term (23)(x+2)(2^3)^{(x+2)}.

    2(x−3)×23(x+2)=272^{(x-3)} \times 2^{3(x+2)} = 2^7 2(x−3)×2(3x+6)=272^{(x-3)} \times 2^{(3x+6)} = 2^7

  4. Apply the exponent rule am×an=am+na^m \times a^n = a^{m+n} to the left side of the equation.

    2(x−3)+(3x+6)=272^{(x-3) + (3x+6)} = 2^7 2(x+3x−3+6)=272^{(x+3x-3+6)} = 2^7 2(4x+3)=272^{(4x+3)} = 2^7

  5. Since the bases are equal, equate the exponents.

    4x+3=74x+3 = 7

  6. Solve the linear equation for xx.

    4x=7−34x = 7 - 3 4x=44x = 4 x=44x = \frac{4}{4} x=1x = 1

    The value of xx is 1\boxed{1}.

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