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Math Help - Log Prob.

  1. #1
    Newbie
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    Log Prob.

    Hi,

    I'm having difficulty beginning to solve the following log problem:

    6^(2log6x + log6x) = 125
    (Where the the two 6's are the base of the logs)

    Any help is much appreciated.

    This is what I have so far:

    6^(log6x^2 + log6x) = 125

    6^(log6 (x^2) (x) ) = 125

    Am I on the right track here? Do I have to make the 6 and 125 the same base to solve?

    Thanks,

    Jim.
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  2. #2
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    Hello, Jim!

    6^{2\log_6x + \log_6x} \:=\: 125

    This is what I have so far:

    6^{\log_6x^2 + \log_6x} \:=\:125

    6^{\log_6(x^2) (x)} \:=\:125

    Am I on the right track here?

    Then we have: . \underbrace{6^{\log_6(x^3)}}_{\downarrow} \:=\:125
    . . . . . . . . . . . . . x^3 \:=\:125 .
    . . . Do you see why?

    . . . . . . . . . . . . . . x \:=\:5

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  3. #3
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    Hi Soroban,

    Thank you for your reply.

    I think I see why:

    is it because 6^(log6) is equivalent to 1?

    so 6^(log6(x^3)) turns into:

    1 (x^3) (b/c
    6^(log6) = 1)

    and then just: x^3

    and so you have:

    x = 5 (b/c the cubed root of 125 is 5)

    is this correct?
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  4. #4
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    Yep, looks good.
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