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Thread: help with this please

  1. #1
    Junior Member
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    help with this please

    not sure what to do
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  2. #2
    Super Member

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    Hello, rj2001!

    Sove for $\displaystyle x\!:\;\;\pi^{1-4x} \:=\:e^{5x} $

    Take logs: .$\displaystyle \ln\left(\pi^{1-4x}\right) \;=\;\ln\left(e^{5x}\right) \quad\Rightarrow\quad(1-4x)\ln(\pi) \;=\;5x\ln(e)$

    Since $\displaystyle \ln(e) = 1$, we have: .$\displaystyle \ln(\pi) - 4x\ln(\pi) \;=\;5x \quad\Rightarrow\quad 4x\ln(\pi) + 5x \;=\;\ln(\pi) $

    Factor: .$\displaystyle x\bigg[4\ln(\pi) + 5\bigg] \;=\;\ln(\pi) $

    Therefore: .$\displaystyle x \;=\;\frac{\ln(\pi)}{4\ln(\pi) + 5} $

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