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Math Help - logarithm problem

  1. #1
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    logarithm problem

    Hi;
    solve logbase6 36^5x.

    6^2 = 36^5x

    2 =5x therefore x=2/5.

    Is that correct.I think it is
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  2. #2
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    Re: logarithm problem

    Or do both sides need to have same base therefore

    6^2 = (6^2)^5x

    giving 6^2 = 6^10x

    resulting in 2 = 10x therefore x = 2/10

    Yeah that looks better.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: logarithm problem

    Presumably, the equation is:

    \log_6(36^{5x})=2

    Converting from logarithmic to exponential form is one way we could proceed, but here is another:

    We may use the log property \log_a(b^c)=c\cdot\log_a(b) to write:

    5x\cdot\log_6(36)=2

    5x\cdot\log_6(6^2)=2

    Using the above property again, we have:

    10x\cdot\log_6(6)=2

    Since \log_a(a)=1 we now have:

    10x=2

    5x=1

    x=\frac{1}{5}
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  4. #4
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    Re: logarithm problem

    Great thank you just needed to make sure.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: logarithm problem

    Another way:

    \log_6(36^{5x})=2=2\cdot\log_6(6)=\log_6(6^2)=\log  _6(36^1)

    Hence:

    5x=1

    x=\frac{1}{5}
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  6. #6
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    Re: logarithm problem

    another way

    6^2= (6^2)^5x
    36 = 36^5x
    5x=1
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