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Math Help - logarithm problem

  1. #1
    Junior Member
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    logarithm problem

    Log3x=y=log9(2x-1)



    where the 3 and 9 is subsript

    i reach to this point
    log3x=log9(2x-1)
    log3x=log3(2x-1)
    log39
    where all the 3 is subscript.
    PS i dont understand how log3(2x-1) is derived from log9(2x-1)
    log39

    where 3 is subscript
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  2. #2
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    re: logarithm problem

    Hello, mathkid12!

    \log_3x \:=\:y\:=\:\log_9(2x-1)

    \text{i reached this point: }\:\log_3x\:=\:\log_9(2x-1)

    . . \log_3x \:=\:\frac{\log_3(2x-1)}{\log_39}

    They used the Base Change Formula: . \log_b(a) \;=\;\frac{\log_n(a)}{\log_n(b)}

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  3. #3
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    re: logarithm problem

    What would be the end answer please and your help is greatly appreciated
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  4. #4
    MHF Contributor Siron's Avatar
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    re: logarithm problem

    To continue on what you had:
    \log_3(x)=\frac{\log_3(2x-1)}{\log_3(9)}
    \Leftrightarrow \log_3(x)=\frac{\log_3(2x-1)}{2}
    \Leftrightarrow 2\log_3(x)=\log_3(2x-1)
    \Leftrightarrow \log_3(x^2)=\log_3(2x-1)

    Now use the fact that \log_a(u(x))=\log_a(v(x)) \Leftrightarrow u(x)=v(x)
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  5. #5
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    re: logarithm problem

    Thanks
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  6. #6
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    Re: logarithm problem

    Note that log_9(x)= y if and only if [tex]x= 9^y= (3^2)^y= 3^{2y}[/itex] which then says that log_3(x)= 2y
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