# Logarithm

• Feb 14th 2011, 07:42 PM
MichaelLight
Logarithm
If log4 N=p and log12 N=q, show that

log
3 48= Attachment 20812

I tried so many times but it seems to be so tough for me... can anyone help?
• Feb 14th 2011, 08:10 PM
pickslides
This might be sending you the long way home but write out $\displaystyle \frac{p+q}{p-q}$ in terms of the logs given then change the base of all logs to base 3.
• Feb 15th 2011, 03:59 AM
HallsofIvy
Quote:

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

log
3 48= Attachment 20812

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?
• Feb 15th 2011, 04:21 AM
MichaelLight
Quote:

Originally Posted by pickslides
This might be sending you the long way home but write out $\displaystyle \frac{p+q}{p-q}$ 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?

Quote:

Originally Posted by HallsofIvy
How did "N" disappear? And where did "48" come from?

I don't know...
• Feb 15th 2011, 05:53 AM
try using $\log_aN=\frac{1}{\log_Na}$ for $\frac{p+q}{p-q}$

EDIT:
forgot to mention

you need $\log_ab=\frac{\log_Nb}{\log_Na}$ as well. That is where "N" disappears.
• Feb 15th 2011, 07:43 AM
Soroban
Hello, MichaelLight!

My approach is worse that Pickslides' suggestion.

I used the Base-Change formula.

Quote:

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

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

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

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

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

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

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