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Thread: exponential and logarithmic

  1. #1
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    Question exponential and logarithmic

    I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?



    log x + log(x -
    3) = 1



    thanks!

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by lizzy326 View Post
    I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?



    log x + log(x -
    3) = 1



    thanks!

    First apply the property $\displaystyle \ln a+\ln b= \ln(ab)$

    We see the left side of the equation becomes $\displaystyle \log(x^2-3x)=1$

    But note that $\displaystyle \log(10)=1$ [I'm assuming that when you refer to log, you mean it is of base 10]

    So we see that $\displaystyle \log(x^2-3x)=\log(10)$

    This is true only when $\displaystyle x^2-3x=10\implies x^2-3x-10=0$

    Solve this quadratic equation and see what x values you get.

    Does this make sense?

    --Chris
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  3. #3
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    Quote Originally Posted by lizzy326 View Post
    I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?



    log x + log(x -
    3) = 1



    thanks!

    What do you mean "when the bases aren't the same"? In this example, you have the same base.

    If you mean "$\displaystyle log_a(x)- log_b(x-3)$ then you have to change logarithm to the same base as the other.

    $\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}$ so
    $\displaystyle log_a(x)- log_b(x-3)= \frac{log_b(x)}{log_b(a)}- log_b(x-3)= \frac{(log_b(x)- log_b(a)log_b(x-1)}{log_b(a)}$
    $\displaystyle = \frac{log_b(x)- log_b((x-1)^{log_b(a)}}{log_b(a)= \frac{log_b(\frac{x}{(x-1)^{log_b(a)})}{log_b(a)}$

    Not very pretty but that's the best you can do.
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