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Math Help - Solving Logarithmic/Exponential Equations

  1. #1
    Newbie radh's Avatar
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    Solving Logarithmic/Exponential Equations

    Hey everyone, I've tried these log problems, but to no avail.

    25^(x-1) = 125^4x
    I tried this:
    (5^2)^x-1 = (5^3)^4x
    5^(2x-2) = 5^12x
    2x-2 = 12x
    Here, I was confused. No substitution would make it work. (The instant math solver says -1/5)
    Am I going at this the wrong way?

    Thank you for the help!
    Last edited by radh; May 11th 2012 at 11:45 PM. Reason: add in parentheses.
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  2. #2
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    Re: Solving Logarithmic/Exponential Equations

    Quote Originally Posted by radh View Post
    Hey everyone, I've tried these log problems, but to no avail.

    25^x-1 = 125^4x
    I tried this:
    (5^2)^x-1 = (5^3)^4x
    5^2x-2 = 5^12x
    2x-2 = 12x
    Here, I was confused. No substitution would make it work. (The instant math solver says -1/5)
    Am I going at this the wrong way?

    Also, for clarification (in two examples), here's another problem:
    36^x-9 = 6^2x
    I tried this:
    (6^2)^x-9 = (6)^2x
    6^2x-18 = 6^2x
    2x-18 = 2x
    Which once again comes up with an impossible substitution.

    Thank you for the help!
    Is it \displaystyle \begin{align*} 25^x - 1 = 125^{4x} \end{align*} or \displaystyle \begin{align*} 25^{x - 1} = 125^{4x} \end{align*}, or something else entirely?
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  3. #3
    Newbie radh's Avatar
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    Re: Solving Logarithmic/Exponential Equations

    It's 25^(x-1) and 36^(x-9). I apologize for the confusion.
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  4. #4
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    Re: Solving Logarithmic/Exponential Equations

    You are correct up to \displaystyle \begin{align*} 2x-2 = 12x \end{align*}, so

    \displaystyle \begin{align*} 2x - 2 &= 12x \\ -2 &= 10x \\ -\frac{2}{10} &= x \\ x &= -\frac{1}{5} \end{align*}
    Thanks from radh
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