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Thread: Problem with changing of base log question

  1. #1
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    Problem with changing of base log question

    Hi all,
    Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

    Heres the question:
    3 log (base 3) x - 26= 9 log (base x) 3

    I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

    Thanks in advance!
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  2. #2
    Super Member Bacterius's Avatar
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    $\displaystyle 3 \log_3 {(x)} - 26= 9 \log_x {(3)}$

    First isolate all the $\displaystyle x$ terms on a side :

    $\displaystyle 3 \log_3 {(x)} - 9 \log_x {(3)}= 26$

    Note that you can factorize out a $\displaystyle 3$ to make it simpler :

    $\displaystyle 3 (\log_3 {(x)} - 3 \log_x {(3)})= 26$

    Divide to simplify :

    $\displaystyle \log_3 {(x)} - 3 \log_x {(3)}= \frac{26}{3}$

    Using the power log law :

    $\displaystyle \log_3 {(x)} - \log_x {(9)}= \frac{26}{3}$

    Now, $\displaystyle \log_x {(9)} = \frac{\log_3 {(9)}}{\log_3 {(x)}}$ (we're using 3 as a changing base for obvious reasons). Substituting :

    $\displaystyle \log_3 {(x)} - \frac{\log_3 {(9)}}{\log_3 {(x)}}= \frac{26}{3}$

    Now, $\displaystyle \log_3 {(9)} = 2$, so :

    $\displaystyle \log_3 {(x)} - \frac{2}{\log_3 {(x)}}= \frac{26}{3}$

    Setting $\displaystyle u = \log_3 {(x)}$, you only have to solve :

    $\displaystyle u - \frac{2}{u} = \frac{26}{3}$

    The solutions are given by substituting another well-chosen value for $\displaystyle u$ so as to find a quadratic equation. The solutions are :

    $\displaystyle u = \frac{1}{3} \left ( 13 \pm \sqrt{187} \right )$

    Now, since $\displaystyle u = \log_3 {(x)}$, $\displaystyle x = 3^u$. Just reverse the substitution to find the values of $\displaystyle x$, and you are done

    Does it make sense ? (hope I'm right, Wolfram won't check it for some reason)
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  3. #3
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    Quote Originally Posted by AeroScizor View Post
    Hi all,
    Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

    Heres the question:
    3 log (base 3) x - 26= 9 log (base x) 3

    I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

    Thanks in advance!
    Is it $\displaystyle 3 \log_3 (x) - 26 = 9 \log_x 3$ or $\displaystyle 3 \log_3 (x - 26) = 9 \log_x 3$?
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  4. #4
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    To Bacterius:
    Great help you have been! Your steps are clear and concise. Thanks for your fast reply!
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Is it $\displaystyle 3 \log_3 (x) - 26 = 9 \log_x 3$ or $\displaystyle 3 \log_3 (x - 26) = 9 \log_x 3$?
    It's the former, not the latter.
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  6. #6
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    Quote Originally Posted by AeroScizor View Post
    Hi all,
    Just doing my revision and came across a logarithms question that involves changing of base. I could only complete halfway before getting stuck. So I hope someone here can help me out.

    Heres the question:
    3 log (base 3) x - 26= 9 log (base x) 3

    I have managed to switch the base of 9 log from x to 3 but soon got stuck at the part whereby 9/log (base 3) x----doesn't matter if you don't understand, just solve it your own way.

    Thanks in advance!
    hi AeroScizor,

    a small correction is needed.

    $\displaystyle 3log_3x-26=9log_x3$

    $\displaystyle 3log_3x=log_x3^9+26$

    $\displaystyle 3log_3x-log_x3^9=26$

    using the change of base formula $\displaystyle log_bx=\frac{log_ax}{log_ab}$ we get

    $\displaystyle 3log_3x-\frac{log_33^9}{log_3x}=26$

    $\displaystyle log_33^9=9$

    $\displaystyle \frac{3\left(log_3x\right)^2-9}{log_3x}=26$

    $\displaystyle 3\left(log_3x\right)^2-9=26log_3x$

    $\displaystyle 3a^2-26a-9=0$

    $\displaystyle a=\frac{26\pm\sqrt{784}}{6}=\frac{26\pm28}{6}=\fra c{54}{6},\ -\frac{2}{6}$

    $\displaystyle log_3x=9,\ -\frac{1}{3}$

    There was a small typo in the earlier post, as $\displaystyle 3log_x3=log_x3^3=log_x27$
    Last edited by Archie Meade; May 10th 2010 at 09:53 AM.
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  7. #7
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    Hello, AeroScizor!

    Solve for $\displaystyle x\!:\;\;3\log_3(x) - 26 \;=\; 9\log_x(3)$
    Formula: .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)} $


    So we have: . $\displaystyle 3\log_3(x) - 26 \;=\;\frac{9}{\log_3(x)} $

    Multiply by $\displaystyle \log_3(x)\!:\quad 3\bigg[\log_3(x)\bigg]^2 - \;26\log_3(x) \;=\;9 \quad\Rightarrow\quad3\bigg[\log_3(x)\bigg]^2 - \;26\log_3(x) \;-\; 9 \;=\;0 $

    Factor: . $\displaystyle \bigg[3\log_3(x) + 1\bigg]\,\bigg[\log_3(x) - 9\bigg] \;=\;0 $


    And we have two equations to solve:

    . . $\displaystyle 3\log_3(x) + 1 \:=\:0 \quad\Rightarrow\quad \log_3(x) \:=\:-\tfrac{1}{3} \quad \hdots\text{ No real roots}$

    . . $\displaystyle \log_3(x) - 9 \:=\:0 \quad\Rightarrow\quad \log_3(x) \:=\:9 \quad\Rightarrow\quad x \:=\:3^9$


    Therefore: .$\displaystyle x \;=\;19,\!683$

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