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Math Help - Proving this

  1. #1
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    Proving this

    Prove that:


    [log_base a_(x)] / [log_base ab_(x)] = 1 + log_base a_(b), for all a, b >1 and x >0.



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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    Prove that:

    [log_base a_(x)] / [log_base ab_(x)] = 1 + log_base a_(b), for all a, b >1 and x > 0.

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    Hello,

    you know that log_a(x) = ln(x)/(lna)

    The LHS of your equation is:

    ((ln(x))/(ln(a)))/((ln(x))/(ln(ab)) = (ln(ab))/(ln(a))=(ln(a)+ln(b))/(ln(a))= 1+(ln(b))/(ln(a)) = 1 + log_a(b)

    EB
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  3. #3
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    Hello, Mr_Green!

    If we're allowed the Base-Change Formula, it's not bad . . .


    Prove that: .[log_a(x)] / [log_ab(x)] .= . 1 + log_a(b)
    We will change the denominator to base a.

    . . . . . . . . .log_a(x) . . . . . .log_a(x)
    We have: . ----------- . = . -------------- . = . log_a(ab)
    . . . . . . . . log_ab(x) . . . . . log_a(x)
    . . . . . . . . . . . . . . . . . . . .-----------
    . . . . . . . . . . . . . . . . . . . .log_a(ab)


    And: . log_a(ab) . = . log_a(a) + log_a(b) . = . 1 + log_a(b)

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