Logarithms

**deltoids** · May 1st 2006, 05:37 PM

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!

**ThePerfectHacker** · May 1st 2006, 06:45 PM

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,
$x=\frac{2\log 5-3\log 4}{\log 4-\log 5}\approx 4.21$

**deltoids** · May 1st 2006, 06:51 PM

Much Appreciated sir!

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