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Math Help - Logarithm help needed please

  1. #1
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    Logarithm help needed please

    Solve each of the following equation to find x in terms of a where a>0 and a not equal to 100. The logarithms are to base 100
    (a) a^x = 10^ 2x+1

    (b) 2 log (2x) = 1 + log a
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Logarithm help needed please

    a) Assuming we have a^x=10^{2x+1}

    Take the \log_{10} of both sides:

    \log_{10}(a^x)=\log_{10}(10^{2x+1})

    Apply the property \log_a(b^c)=c\cdot\log_a(b):

    x\cdot\log_{10}(a)=(2x+1)\cdot\log_{10}(10)

    On the right, apply the property \log_a(a)=1:

    x\cdot\log_{10}(a)=2x+1

    Solve for x.

    b) 2\log_{100}(2x)=1+\log_{100}(a)

    \log_{100}((2x)^2)=\log_{100}(100)+\log_{100}(a)

    On the right, apply the property \log_a(b)+\log_a(c)=\log_a(bc):

    \log_{100}(4x^2)=\log_{100}(100a)

    Equate arguments, and take the valid root.
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  3. #3
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    Re: Logarithm help needed please

    Hello, mysur!

    Are you sure of the instructions?


    Solve each of the following equations to find x in terms of a
    where a>0 and a \ne 100. .The logarithms are to base 100. . Really?

    (b)\;2\log_{100}(2x) \:=\: 1 + \log_{100}(a)

    We have: . 2\log_{100}(2x) \;=\;1 + \log_{100}(a)

    . . . . . . . . 2\log_{100}(2x) \;=\;\log_{100}(100) + \log_{100}(a)

    . . . . . . . . 2\log_{100}(2x) \;=\;\log_{100}(100a)

    . . . . . . . . . \log_{100}(2x) \;=\;\tfrac{1}{2}\log_{100}(100a)

    . . . . . . . . . \log_{100}{(2x) \;=\;\log_{100}(100a)^{\frac{1}{2}}

    . . . . . . . . . \log_{100}(2x) \;=\;\log_{100}(10\sqrt{a})

    I . . . . . . . . . . . . . 2x \;=\;10\sqrt{a}

    . . . . . . . . . . . . . . . x \;=\;5\sqrt{a}
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