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Math Help - Solving Exponential Equation different base(non common)

  1. #1
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    Solving Exponential Equation different base(non common)

    Here's the problem:
    2^{2x+12}= 3^{x-34}

    I can get it to this by taking the log of both sides:
    \frac{2x+12}{x-34}=\frac{log(3)}{log(2)}

    Now I could just use a calculator and get the value for the right side and the clean it up. But how would I finish this problem and get the answer in terms of log or ln. Can someone give me a hand? Thanks!
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  2. #2
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    Re: Solving Exponential Equation different base(non common)

    Just cross multiply and collect terms containing x on one side. You will have
    2x log 2 + 12 log 2 = x log 3 - 34 log 3
    x log 4 + 12 log 2 = x log 3 - 34 log 3
    x log 4 - x log 3 = - 12 log 2- 34 log 3
    x log 4/3 = - 12 log 2- 34 log 3 etc.
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  3. #3
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    Re: Solving Exponential Equation different base(non common)

    Hello, jjtjp!

    \text{Solve for }x\!:\;\;2^{2x+12}\:=\: 3^{x-34}

    I can get it to this by taking the log of both sides:

    \frac{2x+12}{x-34}=\frac{\log(3)}{\log(2)}
    This is correct, but why did you create fractions?

    We have: . 2^{2x+12} \:=\:3^{x-34}

    Take logs: . \ln\left(2^{2x+12}\right) \:=\:\ln\left(3^{x-34}\right)

    . . . . . . . (2x+12)\ln2 \:=\:(x-34)\ln3

    n . . . . 2x\ln2 + 12\ln2 \:=\:x\ln3 - 34\ln3

    . . . . . . 2x\ln2 - x\ln3 \:=\:-34\ln3 - 12\ln 2

    . . . . . . x(2\ln2 - \ln3) \:=\:-(34\ln3 + 12\ln2)

    . . . . . . . . . . . . . . . x \:=\:-\frac{34\ln3 + 12\ln2}{2\ln2-\ln3}

    . . . . . . . . . . . . . . . x \:\approx\: -158.7536672
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    Re: Solving Exponential Equation different base(non common)

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