# Math Help - Logarithms

1. ## Logarithms

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

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

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

I would think that the setup involves something with logs, but that's about all I know. Thanks in advance!
You have,
$4^{x+3}=5^{x+2}$
Take the log of both sides,
$\log 4^{x+3}=\log 5^{x+2}$
Thus, by exponent rules,
$(x+3)\log 4=(x+2)\log 5$
Thus,
$x\log 4+3\log 4=x\log 5+2\log 5$
Thus,
$x\log 4-x\log 5=2\log 5-3\log 4$
Thus,
$x(\log 4-\log 5)=2\log 5-3\log 4$
Thus,
$x=\frac{2\log 5-3\log 4}{\log 4-\log 5}\approx 4.21$

3. Much Appreciated sir!