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Thread: Logarithm problem (sorry if this in the wrong section!)

  1. #1
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    Logarithm problem (sorry if this in the wrong section!)

    I couldn't seem to find a board on logarithms so apologised if this is in the incorrect section. Apologies for formatting too.

    I'm trying to work out log 4^2x-1 = log 3^x+1

    I've done this so far (but I think I've made it more difficult than it needs to be)

    i) (2x 1)log 4 = (x + 1)log 3
    ii) 2x-1= (x+1)log⁡3 / log⁡4
    iii) 2x - (x+1)log 3 / log 4 = 1
    but I do not know where to go from there...and like I said I think there is an easier more straightforward way to do this.

    I'd appreciate any help. Thank you.
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle \begin{gathered}
    \left( {2x - 1} \right)\log 4 = \left( {x + 1} \right)\log 3 \hfill \\
    \leftrightarrow \left( {2\log 4 - \log 3} \right)x = \log 3 + \log 4 \hfill \\
    \leftrightarrow x = \frac{{\log 12}}
    {{\log \frac{{16}}
    {3}}} \hfill \\
    \end{gathered}
    $
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  3. #3
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    Hello, Jamie!

    Solve: . $\displaystyle \log\left(4^{2x-1}\right) \: = \:\log\left(3^{x+1}\right)$

    I've done this so far . . .

    $\displaystyle (2x 1)\log 4 \:= \x + 1)\log 3$ . . . . Right!
    We want to solve for $\displaystyle x$, right?


    Expand: . $\displaystyle 2x\!\cdot\!\log 4 - \log4 \;=\;x\!\cdot\!\log3 + \log 3$

    Get the $\displaystyle x$'s together: . $\displaystyle 2x\!\cdot\log4 - x\!\cdot\!\log3 \;=\;\log 3 + \log 4$

    Factor: . $\displaystyle x\left(2\log4 - \log 3\right) \:=\:\log 3 + \log 4$

    Therefore: . $\displaystyle \boxed{x \;=\;\frac{\log3 + \log4}{2\log4 - \log3}} $


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Peritus simplified it further, using logarithm rules.

    . . $\displaystyle \log3 + \log4 \;=\;\log(3\cdot4) \;=\;\log12$

    . . $\displaystyle 2\log4 - \log 3 \;=\;\log(4^2) - \log3 \;=\;\log16 - \log3 \;=\;\log\left(\frac{16}{3}\right) $

    And we have: . $\displaystyle \frac{\log(12)}{\log\left(\frac{16}{3}\right)} \;=\;\log_{\frac{16}{3}}(12)$


    I've always wanted to do that!
    .
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