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!

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- May 1st 2006, 05:37 PMdeltoidsLogarithms
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 PMThePerfectHackerQuote:

Originally Posted by**deltoids**

$\displaystyle 4^{x+3}=5^{x+2}$

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 PMdeltoids
Much Appreciated sir!

http://img.photobucket.com/albums/v5...iticool/ty.jpg