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Thread: 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 $\displaystyle a^x=10^{2x+1}$

    Take the $\displaystyle \log_{10}$ of both sides:

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

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

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

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

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

    Solve for $\displaystyle x$.

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

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

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

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

    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 $\displaystyle x$ in terms of $\displaystyle a$
    where $\displaystyle a>0$ and $\displaystyle a \ne 100.$ .The logarithms are to base 100. . Really?

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

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

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

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

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

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

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

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

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