1. ## Logarithm

If log4 N=p and log12 N=q, show that

log
3 48=

I tried so many times but it seems to be so tough for me... can anyone help?

2. This might be sending you the long way home but write out $\displaystyle \displaystyle \frac{p+q}{p-q}$ in terms of the logs given then change the base of all logs to base 3.

3. Originally Posted by MichaelLight
If log4 N=p and log12 N=q, show that

log
3 48=

I tried so many times but it seems to be so tough for me... can anyone help?
How did "N" disappear? And where did "48" come from?

4. Originally Posted by pickslides
This might be sending you the long way home but write out $\displaystyle \displaystyle \frac{p+q}{p-q}$ in terms of the logs given then change the base of all logs to base 3.
I tried but i couldn't get the required answer, have you tried it before?

Originally Posted by HallsofIvy
How did "N" disappear? And where did "48" come from?
I don't know...

5. try using $\displaystyle \log_aN=\frac{1}{\log_Na}$ for $\displaystyle \frac{p+q}{p-q}$

EDIT:
forgot to mention

you need $\displaystyle \log_ab=\frac{\log_Nb}{\log_Na}$ as well. That is where "N" disappears.

6. Hello, MichaelLight!

My approach is worse that Pickslides' suggestion.

I used the Base-Change formula.

$\displaystyle \text{If }\,\log_4 N=p\,\text{ and }\,\log_{12}N=q,\,\text{ show that: }\:\log_3 48\:=\: \dfrac{p+q}{p-q}$

$\displaystyle \displaystyle\text{We have: }\:\log_348 \;\;=\;\;\frac{\ln48}{\ln3} \;\;=\;\;\frac{\ln(12\cdot4)}{\ln(\frac{12}{4})} \;\;=\;\;\frac{\ln12 + \ln4}{\ln12 - \ln4}$

Divide numerator and denominator by $\displaystyle (\ln12\!\cdot\!\ln4)\!:$

. . $\displaystyle \displaystyle =\;\;\frac{\;\dfrac{\ln12 + \ln4}{\ln12\cdot\ln4}\;} {\dfrac{\ln12 - \ln4}{\ln12\cdot\ln4}} \;\;=\;\;\frac{\;\dfrac{1}{\ln4} + \dfrac{1}{\ln12}\;}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}}$

Multiply by $\displaystyle \dfrac{\ln N}{\ln N}\!:$

. . $\displaystyle \displaystyle =\;\;\frac{\ln N}{\ln N}\cdot\frac{\dfrac{1}{\ln4} + \dfrac{1}{\ln12}}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}} \;\;=\;\;\frac{\;\dfrac{\ln N}{\ln 4} + \dfrac{\ln N}{\ln12}\;}{\dfrac{\ln N}{\ln 4} - \dfrac{\ln N}{\ln 12}}$

. . $\displaystyle \displaystyle =\;\;\frac{\log_4\!N + \log_{12}\!N}{\log_4\!N - \log_{12}\!N}} \;\;=\;\;\frac{p + q}{p - q}$