• Apr 9th 2011, 08:02 AM
mathguy80
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
$
log_{\sqrt{12}} 4 + log_{12} \sqrt{4}
$

$
= 2log_{\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}
$

$
\frac{5}{2}log_{12} 4
$

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?
• Apr 9th 2011, 09:00 AM
Soroban
Hello, mathguy80!

Quote:

I am able to solve them okay most of the time.

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
$

. . 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 $\log(2)\!:$

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

• Apr 9th 2011, 10:59 AM
Scurmicurv
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.
• Apr 9th 2011, 11:16 AM
veileen
O.O

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

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 $a\neq 1, b\neq 1$.
• Apr 9th 2011, 01:40 PM
Soroban
Quote:

Originally Posted by veileen

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

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

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

. . Is this relavant?

• Apr 9th 2011, 01:44 PM
veileen
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.
• Apr 9th 2011, 06:46 PM
mathguy80
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!