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Math Help - Easy Proof

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

    Prove that 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 \frac{d}{dx}(log_ax)=\frac{1}{(lna)x}, they rewrite log_ax as \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 \log_a{x}=\frac{\ln x}{\ln a}.

    Let \log_a{x} = y.

    Write in exponential form:
    x = a^y

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

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

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

    Substitute \log_a{x} back in for y:
    \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 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 \frac{d}{dx}(log_ax)=\frac{1}{(lna)x}, they rewrite log_ax as \frac{lnx}{lna}. I need to understand this so that I may sleep tonight. Thanx.
    If x = b^a then a = \log_b{x}.

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

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


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

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

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


    Thus \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 \log_a{x} back in for y:
    \frac{\ln x}{\ln a} = \log_a{x}

    Done!


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    Very Good! Unquestionably proven!
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