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Math Help - logarithmic help

  1. #1
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    logarithmic help

    Use the relationship x = a^y ? y = loga x to solve each of the following:
    log5 25 = x

    log3 x = 0

    How would I go about doing those? I believe the 5 and 3 are the base.
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  2. #2
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    log_{5}25=x

    5^x=25 (\because log_{a}b=c\iff a^c=b)

    5^x=5^2

    \therefore x=2 (comparing exponents)



    log_{3}0=x

    3^x=0
    Which is impossible, so, no solution.
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  3. #3
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    Quote Originally Posted by Referos View Post
    log_{5}25=x

    5^x=25 (\because log_{a}b=c\iff a^c=b)

    5^x=5^2

    \therefore x=2 (comparing exponents)
    Exactly right.


    log_{3}0=x

    3^x=0
    Which is impossible, so, no solution.
    No, the problem was log_3 x= 0 which is the same as x= 3^0= 1.
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  4. #4
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    D'oh! Of course, you are correct; I misread the equation.
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