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Math Help - Question about logarithms

  1. #1
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    Question about logarithms

    Hi Guys,

    I having some trouble with logarithms again. I am able to solve them okay most of the time. But my final answer answer doesn't match the given one.

    Take the following problem for instance. I solved upto
    <br />
log_{\sqrt{12}} 4 + log_{12} \sqrt{4} <br />

    <br />
= 2log_{\sqrt{12}} 2 + log_{12} 2<br />

     <br />
= \dfrac{2}{log_2 \sqrt{12}} + \dfrac{1}{log_2 12}<br />

    <br />
= \dfrac{2}{\frac{1}{2} {log_2 12}} + \dfrac{1}{log_2 12}<br />

    <br />
= \dfrac{4}{log_2 12} + \dfrac{1}{log_2 12}<br />

    <br />
= \dfrac{5}{log_2 12}<br />

    The required answer is,

    <br />
\frac{5}{2}log_{12} 4<br />

    In this case I see that I could get to the required answer by going flipping the base, 4^1/2 and so on.

    In this example I am close to the final answer, but many times I am not this close. Is there a general rule to follow when simplifying? How do I figure out what base the answer should be in? Where should the simplification end?
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  2. #2
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    Hello, mathguy80!

    I am able to solve them okay most of the time.
    But my final answer answer doesn't match the given one.

    I solved up to:

     \log_{\sqrt{12}} 4 + \log_{12} \sqrt{4} \;=\;= 2\log_{\sqrt{12}} 2 + \log_{12} 2

    . .  = \;\dfrac{2}{\log_2 \sqrt{12}} + \dfrac{1}{\log_2 12} \;=\; \dfrac{2}{\frac{1}{2} {\log_2 12}} + \dfrac{1}{\log_2 12}

    . . =\; \dfrac{4}{\log_2 12} + \dfrac{1}{\log_2 12} \;=\; \dfrac{5}{\log_2 12}

    The required answer is: . \frac{5}{2}\log_{12} 4<br />

    Your work is excellent . . . and your answer is correct!

    Their answer is completely arbitrary.
    . . both answers can be written in a number of ways.


    Your answer: . \dfrac{5}{\log_2(12)} \;=\;5\log_{12}(2) \;=\;\dfrac{5\log(2)}{\log(12)} \;=\;\cdots

    Their answer: . \dfrac{5}{2}\log_{12}(4) \;=\;\dfrac{5}{2}\cdot\dfrac{1}{\log_4(12)} \;=\;\dfrac{5\log(4)}{2\log(12)} \;=\; \cdots


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


    I can take their answer and simplfy it beyond all recognition . . .


    \dfrac{5}{2}\log_{12}(4) \;\;=\;\;\dfrac{5\log(4)}{2\log(12)} \;\;=\;\;\dfrac{5\log(2^2)}{2\log(2^2\cdot3)}

    . . . . . . . . =\;\;\dfrac{10\log(2)}{2\left[\log(2^2) + \log(3)\right]} \;\;=\;\;\dfrac{5\log(2)}{2\log(2) + \log(3)}


    Divide numerator and denominator by \log(2)\!:

    . . . . \dfrac{5}{2 + \frac{\log(3)}{\log(2)}} \;\;=\;\;\dfrac{5}{2 + \log_2(3)} \quad \hdots \text{ See?}

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  3. #3
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    It's really as much a question of style as anything else at this point. Though I have noticed that quite a lot people don't seem to like having logarithms in a denominator, just as many prefer not having a root expression there. I've never really heard a good explanation for it and don't care much about it myself.
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  4. #4
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    O.O
    "Their answer is completely arbitrary." No, it's not, your answer seems completely arbitrary.

    \frac{5}{2} \cdot log_{12} 4=\frac{5}{2} \cdot 2 \cdot log_{12} 2=5 \cdot log_{12} 2=\frac{5}{log_2 12}

    I used: log_{a} b=\frac{1}{log_{b} a} and log_{a} b^n=n \cdot log_{a} b, where a, b>0 and a\neq 1, b\neq 1.
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  5. #5
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    Quote Originally Posted by veileen View Post
    "Their answer is completely arbitrary."
    No, it's not, your answer seems completely arbitrary.

    \frac{5}{2}\!\cdot\!\log_{12}4 \:=\:\frac{5}{2}\!\cdot\!2\!\cdot\!\log_{12} 2\:=\:5\!\cdot\!\log_{12}2\:=\:\frac{5}{\log_2 12}

    \text{I used: }\:\log_{a} b\,=\,\frac{1}{\log_{b} a}\,\text{ and }\,\log_{a} b^n\,=\,n\!\cdot\!\log_{a}b,
    . . \text{where }a, b>0 \text{ and }a\neq 1,\:b\neq 1.

    Exactly what is your point?

    OF COURSE my answer is arbitrary!
    I simplifed it to extremes to prove a point.


    Are you claiming that their answer is better?
    You didn't get it either ... just the same answer as mathguy80.


    You quoted some log rules . . . so?

    I used: . \begin{Bmatrix}\log_ba \:=\:\dfrac{\log a}{\log b} \\ \\[-3mm]<br />
\log_b(xy) \:=\:\log_bx + \log_by \\ \\[-3mm]<br />
\log_b(x^n) \:=\:n\!\cdot\!\log_bx \end{Bmatrix}

    . . Is this relavant?

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  6. #6
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    I know is the same thing -_-

    When you do this kind of exercises the point is to obtain the simplest result and this is their result.
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  7. #7
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    Thanks @Soroban. That helps a lot! I kept thinking my reasoning was incorrect, everytime I went to check my answers. It's good to know that I was on the right track all along!

    Thanks @Scurmicurv for the tip about not having the log in the denominator, I think I will try to do that now onwards.

    I learned something new again. Thanks everyone!
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