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Thread: Variable in exponent

  1. #1
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    Variable in exponent

    Hi!

    Solve $\displaystyle e^{2x} = 3^{x - 4}$

    I have used $\displaystyle \log_3$ but this is where it leads:

    $\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

    $\displaystyle 2x \log_3 e = x - 4$

    $\displaystyle 2x \log_3 e + 4 = x$

    $\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac12$
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Variable in exponent

    Quote Originally Posted by Unreal View Post
    Hi!

    Solve $\displaystyle e^{2x} = 3^{x - 4}$

    I have used $\displaystyle \log_3$ but this is where it leads:

    $\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

    $\displaystyle 2x \log_3 e = x - 4$

    $\displaystyle 2x \log_3 e + 4 = x$

    $\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac12$
    What happened in that last line?

    $\displaystyle 2x \log_3 e + 4 = x$

    $\displaystyle 2x \log_3 e - x = -4$

    Now factor the x on the LHS and solve.

    -Dan
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  3. #3
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    Re: Variable in exponent

    Quote Originally Posted by Unreal View Post
    Hi!

    Solve $\displaystyle e^{2x} = 3^{x - 4}$

    I have used $\displaystyle \log_3$ but this is where it leads:

    $\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

    $\displaystyle 2x \log_3 e = x - 4$

    $\displaystyle 2x \log_3 e + 4 = x$
    This is correct up to here although I think it would have been better to take the logarithm with respect to e, not 3, getting
    2x= (x+ 3) ln(3) just because you can easily find a numerical value for log(3). But, as I say, what you have done is correct to this point.

    $\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac{1}{2}$
    But here you have gone wrong. Dividing the left side by "2x" leaves $\displaystyle log_3(e)+ \frac{2}{x}= \frac{1}{2}$

    Subtract $\displaystyle log_3(e)$ from both sides to get $\displaystyle \frac{2}{x}= \frac{1}{2}- log_3(x)$ and then, since "x" is in the denominator on the left, take the reciprocal of both sides:
    $\displaystyle \frac{x}{2}= \frac{1}{\frac{1}{2}- log_2(x)}$

    $\displaystyle x= \frac{2}{\frac{1}{2}- log_3(e)}$
    Last edited by HallsofIvy; Apr 29th 2013 at 07:42 AM.
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