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Thread: Easy Proof

  1. #1
    No one in Particular VonNemo19's Avatar
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    Easy Proof

    Prove that $\displaystyle log_ax=\frac{lnx}{lna}$

    The reason I'm asking this is because I was sleeping through logs. I'm now in calculus and in the proof of $\displaystyle \frac{d}{dx}(log_ax)=\frac{1}{(lna)x}$, they rewrite $\displaystyle log_ax$ as $\displaystyle \frac{lnx}{lna}$. I need to understand this so that I may sleep tonight. Thanx.
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  2. #2
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    Prove that $\displaystyle \log_a{x}=\frac{\ln x}{\ln a}$.

    Let $\displaystyle \log_a{x} = y$.

    Write in exponential form:
    $\displaystyle x = a^y$

    Solve for y, but using natural logs. We do this by taking the natural log of both sides:
    $\displaystyle \ln x = \ln (a^y)$

    Use the property of logs to move the y over:
    $\displaystyle \ln x = y \ln a$

    Divide both sides by $\displaystyle \ln a$:
    $\displaystyle \frac{\ln x}{\ln a} = y$

    Substitute $\displaystyle \log_a{x}$ back in for y:
    $\displaystyle \frac{\ln x}{\ln a} = \log_a{x}$

    Done!


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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Prove that $\displaystyle log_ax=\frac{lnx}{lna}$

    The reason I'm asking this is because I was sleeping through logs. I'm now in calculus and in the proof of $\displaystyle \frac{d}{dx}(log_ax)=\frac{1}{(lna)x}$, they rewrite $\displaystyle log_ax$ as $\displaystyle \frac{lnx}{lna}$. I need to understand this so that I may sleep tonight. Thanx.
    If $\displaystyle x = b^a$ then $\displaystyle a = \log_b{x}$.

    But we can evaluate $\displaystyle a$ in another way, using the formula

    $\displaystyle \log{x^p} = p\log{x}$.


    So $\displaystyle \log{x} = \log{b^a}$

    $\displaystyle \log{x} = a\log{b}$

    $\displaystyle a = \frac{\log{x}}{\log{b}}$.


    Thus $\displaystyle \log_b{x} = \frac{\log{x}}{\log{b}}$.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yeongil View Post
    Substitute $\displaystyle \log_a{x}$ back in for y:
    $\displaystyle \frac{\ln x}{\ln a} = \log_a{x}$

    Done!


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    Very Good! Unquestionably proven!
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