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Thread: graphing logarithmic equation

  1. #1
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    graphing logarithmic equation

    How do you graph this?

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  2. #2
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    lo{g_n}({(mn)^{x - {{\log }_m}n}}) = {\log _n}(\frac{{{{(mn)}^x}}}{{{{(mn)}^{{{\log }_m}n}}}}) = {\log _n}(\frac{{{m^x}{n^x}}}{{{m^{{{\log }_m}n}} \times {n^{{{\log }_m}n}}}}) = <br />
     = {\log _n}(\frac{{{m^x}{n^x}}}{{n \times {n^{{{\log }_m}n}}}}) = {\log _n}({n^{x - 1}}) + {\log _n}(\frac{{{m^x}}}{{{n^{{{\log }_m}n}}}}) = (x - 1) + x{\log _n}m - <br />
     - {\log _n}({n^{{{\log }_m}n}}) = x - 1 + x{\log _n}m - {\log _m}n = x - 1 + x{\log _n}m - \frac{1}{{{{\log }_n}m}} = <br />
     = ({\log _n}m + 1)x + ( - 1 - \frac{1}{{{{\log }_n}m}})<br />
    This is line (f(x)=ax+b). Can you graph this now?
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  3. #3
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    Hello, Anemori!

    Slightly different approach . . . same answer.


    How do you graph this?

    . . f(x) \;=\;\log_n(mn)^{x-\log_mn}\,\;\;m,n > 0

    f(x) \;=\;\log_n\bigg[m^{x-\log_mn}\cdot n^{x-\log_mn}\bigg]

    . . . =\;\log_n\left(m^{x-\log_mn}\right) + \log_n\left(n^{x-\log_mn}\right)

    . . . =\; (x-\log_mn)\log_nm + (x-\log_mn)\underbrace{\log_nn}_{\text{This is 1}}

    . . . =\; (\log_nm)x - \underbrace{(\log_mn)(\log_nm)}_{\text{This is 1}} + x - \log_mn

    f(x) \;=\; \left(1 + \log_nm\right)x - \left(1 + \log_mn\right)


    We have: . f(x) \:=\:mx + b\quad\hdots\;\text{ a line}

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  4. #4
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    <--- this is Icanfly answer.


    and this is Soroban answer. which one is right?


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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Anemori!

    Slightly different approach . . . same answer.



    f(x) \;=\;\log_n\bigg[m^{x-\log_mn}\cdot n^{x-\log_mn}\bigg]

    . . . =\;\log_n\left(m^{x-\log_mn}\right) + \log_n\left(n^{x-\log_mn}\right)

    . . . =\; (x-\log_mn)\log_nm + (x-\log_mn)\underbrace{\log_nn}_{\text{This is 1}}

    . . . =\; (\log_nm)x - \underbrace{(\log_mn)(\log_nm)}_{\text{This is 1}} + x - \log_mn

    f(x) \;=\; \left(1 + \log_nm\right)x - \left(1 + \log_mn\right)


    We have: . f(x) \:=\:mx + b\quad\hdots\;\text{ a line}


    How can I find points and y-interecept and x-intercept on this equation?
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  6. #6
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    Ok the original problem is how do you graph this and finding intercepts?
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  7. #7
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    Quote Originally Posted by Anemori View Post
    Ok the original problem is how do you graph this and finding intercepts?
    they are both correct , check the properties of logarithms .

    As for the x-intercept , when f(x)=0 , x=...

    y-intercept , when x=0 , f(0)=...

    calculate that
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