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Math Help - Logs with divided base

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    Logs with divided base

    How do you solve a problem that has a divided variable as a base of a logarithm? My problem is Loga(x)/Log a/b(x). I guess that would be read; Log 'a' of x divided by Log 'a/b' of x. Thanks
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    Re: Logs with divided base

    Quote Originally Posted by ThanksForHelp View Post
    How do you solve a problem that has a divided variable as a base of a logarithm? My problem is Loga(x)/Log a/b(x). I guess that would be read; Log 'a' of x divided by Log 'a/b' of x. Thanks

    What is your original problem?
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    Re: Logs with divided base

    Quote Originally Posted by ThanksForHelp View Post
    How do you solve a problem that has a divided variable as a base of a logarithm? My problem is Loga(x)/Log a/b(x). I guess that would be read; Log 'a' of x divided by Log 'a/b' of x. Thanks
    In loga(x), which one is base? a or x? If a is the base, use change of base rule. You can take any base. So the problem becomes

    Loga(x)/Log a/b(x) = [log(x)/log(a)]/[log(x)/log(a/b)] = log(a/b)/log(a) = [log(a) - log(b)]/log(a) = 1 - log(b)/log(a) = 1 - loga(b) i.e 1 - log b to the base a.
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    Re: Logs with divided base

    Quote Originally Posted by ThanksForHelp View Post
    How do you solve a problem that has a divided variable as a base of a logarithm? My problem is Loga(x)/Log a/b(x). I guess that would be read; Log 'a' of x divided by Log 'a/b' of x. Thanks
    Dear ThanksForHelp,

    Let, \log_{\frac{a}{b}}x=y

    \left(\frac{a}{b}\right)^y=x

    a^y=b^{y}x

    y=y\log_{a}b+\log_{a}x

    y=\frac{\log_{a}x}{1-\log_{a}b}

    Therefore, \log_{\frac{a}{b}}x=\frac{\log_{a}x}{1-\log_{a}b}

    I hope this will help you.
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    Re: Logs with divided base

    Quote Originally Posted by sa-ri-ga-ma View Post
    In loga(x), which one is base? a or x? If a is the base, use change of base rule. You can take any base. So the problem becomes

    Loga(x)/Log a/b(x) = [log(x)/log(a)]/[log(x)/log(a/b)] = log(a/b)/log(a) = [log(a) - log(b)]/log(a) = 1 - log(b)/log(a) = 1 - loga(b) i.e 1 - log b to the base a.
    Thanks that was perfect! I knew to change the base, but was thrown by the fact that the base was a divided variable. Thank you sooooo much. Extremely helpful.
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    Re: Logs with divided base

    Thanks for everyone's time.
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    Re: Logs with divided base

    Hello, ThanksForHelp!

    Are you familiar with the Base-Change Formula?

    . . \log_b(x) \:=\:\frac{\ln x}{\ln b}


    \text{Simplify: }\:Y \;=\; \frac{\log_a(x)}{\log_{\frac{a}{b}}(x)}

    We have: . \begin{Bmatrix}\log_a(x) &=& \dfrac{\ln x}{\ln a} \\ \\[-3mm] \log_{\frac{a}{b}}(x) &=& \dfrac{\ln x}{\ln\frac{a}{b}} \end{Bmatrix}

    \text{Then: }\;Y \;=\;\frac{\,\dfrac{\ln x}{\ln a}\,}{\dfrac{\ln x}{\ln\frac{a}{b}}} \;=\;\frac{\ln x}{\ln a}\cdot\frac{\ln\frac{a}{b}}{\ln x}} \;=\;\frac{\ln\frac{a}{b}}{\ln a}

    \text{Therefore: }\;Y \;=\;\frac{\ln a - \ln b}{\ln a}\;\text{ or }\;1 - \frac{\ln b}{\ln a}

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