# Logarithms

• May 1st 2006, 05:37 PM
deltoids
Logarithms
Final exam's tomorrow, and I have a lot of trouble with this type of a problem:

http://img326.imageshack.us/img326/4870/554nv.jpg

I would think that the setup involves something with logs, but that's about all I know. Thanks in advance!
• May 1st 2006, 06:45 PM
ThePerfectHacker
Quote:

Originally Posted by deltoids
Final exam's tomorrow, and I have a lot of trouble with this type of a problem:

http://img326.imageshack.us/img326/4870/554nv.jpg

I would think that the setup involves something with logs, but that's about all I know. Thanks in advance!

You have,
$\displaystyle 4^{x+3}=5^{x+2}$
Take the log of both sides,
$\displaystyle \log 4^{x+3}=\log 5^{x+2}$
Thus, by exponent rules,
$\displaystyle (x+3)\log 4=(x+2)\log 5$
Thus,
$\displaystyle x\log 4+3\log 4=x\log 5+2\log 5$
Thus,
$\displaystyle x\log 4-x\log 5=2\log 5-3\log 4$
Thus,
$\displaystyle x(\log 4-\log 5)=2\log 5-3\log 4$
Thus,
$\displaystyle x=\frac{2\log 5-3\log 4}{\log 4-\log 5}\approx 4.21$
• May 1st 2006, 06:51 PM
deltoids