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Math Help - Solving for x with logs

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    Solving for x with logs

    Problem: 2^2x=k

    Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
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    Re: Solving for x with logs

    Quote Originally Posted by benny92000 View Post
    Problem: 2^2x=k

    Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
    Either

    \displaystyle \begin{align*} 2^{2x} &= k \\ 2x &= \log_2{k} \\ x &= \frac{1}{2}\log_2{k} \end{align*}

    or

    \displaystyle \begin{align*} 2^{2x} &= k \\ \ln{\left(2^{2x}\right)} &= \ln{k} \\ 2x\ln{2} &= \ln{k} \\ 2x &= \frac{\ln{k}}{\ln{2}} \\ x &= \frac{\ln{k}}{2\ln{2}} \end{align*}
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    Re: Solving for x with logs

    Quote Originally Posted by benny92000 View Post
    Problem: 2^2x=k

    Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
    2^{2x} = k

    \log(2^{2x}) = \log{k}

    2x\log{2} = \log{k}

    x = \frac{\log{k}}{2\log{2}}
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    Re: Solving for x with logs

    Can I also start by taking the log2 of both sides?
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    Re: Solving for x with logs

    Quote Originally Posted by benny92000 View Post
    Can I also start by taking the log2 of both sides?
    You obviously didn't read my post properly, as that was the first method I showed...
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    Re: Solving for x with logs

    Quote Originally Posted by benny92000 View Post
    Problem: 2^2x=k

    Maybe someone can tell me what to look for. I am going to fiddle with it a little more.

    Start by taking the natural log of both sides of the equation.

    Let ln = natural log

    2^(2x) = k

    ln[2^(2x)] = ln(k)

    When we take the natural log of both sides, exponents come down infront of the ln symbol.


    So, 2x comes down and it now looks like this:


    2xln(2) = ln(k)

    At this point, solve the equation for x, dividing both sides by 2(ln2).

    x = ln(k)/2(ln2)
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    Re: Solving for x with logs

    Quote Originally Posted by benny92000 View Post
    Can I also start by taking the log2 of both sides?
    you can use any log base you want.

    I believe this thread has been beat to death.
    Last edited by skeeter; November 29th 2011 at 03:24 PM.
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