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Math Help - Question with Logarithms

  1. #1
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    Question with Logarithms

    How do I turn 3^x = 4^x^-^2 into \frac{\log\frac1{16}}{log\frac3{4}} and could you please also explain the steps thoroughly. Thanks!
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  2. #2
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    Re: Question with Logarithms

    1st, you don't. The first is an EQUATION and the second is an EXPRESSION. Not the same at all.

    2nd, this is a logarithm problem, as you can see by the final form. If everything is suitable, then a^{b} = c^{d} implies b\cdot log(a) = d\cdot log(c). That's almost all you need.
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    Re: Question with Logarithms

    OH sorry i completely forgot to make the second equal to x.
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    Re: Question with Logarithms

    Quote Originally Posted by Jayissocool View Post
    How do I turn 3^x = 4^x^-^2 into \frac{\log\frac1{16}}{log\frac3{4}} and could you please also explain the steps thoroughly. Thanks!
    I'll show you how to start and leave the rest for you:

    3^x = 4^{x-2}

    3^x = 4^x \cdot 4^{-2}

    \frac{3^x}{4^x}=\frac1{16}

    \left(\frac34 \right)^x=\frac1{16}

    ... and now it's your turn.
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  5. #5
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    Re: Question with Logarithms

    Hello, Jayissocool!

    First of all, that is a really stupid way to leave the answer!


    \text{How do I turn }\,3^x \:=\:4^x^{-2}\,\text{ into }\,x \:=\:\frac{\log\frac1{16}}{\log\frac3{4}}\;?

    Take logs: . \log\left(3^x) \;=\;\log\left(4^{x-2}\right) \quad\Rightarrow\quad x\log3 \;=\;(x-2)\log4

    . . x\log 3 \;=\;x\log4 - 2\log 4 \quad\Rightarrow\quad x\log 3 - x\log4 \;=\;-2\log 4

    . . x(\log3-\log4) \;=\;\log(4^{-2}) \quad\Rightarrow\quad x \;=\;\frac{\log(4^{-2})}{\log3-\log 4}

    Therefore: . x \;=\;\dfrac{\log\frac{1}{16}}{\log\frac{3}{4}} \;\;\hdots\text{ which equals: }\,x \,=\,\frac{\log16}{\log\frac{4}{3}}

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