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confusedbuzz · Apr 15, 2008, 07:45 AM

Can someone explain to be the concepts and procedure to solving something like the following question?
(I) log4 2 ! Log9 (2x ! 5) # log8 64

confusedbuzz · Apr 15, 2008, 07:47 AM

Ugh that didn't come out right. Anyway, all numbers directly after the small boxes are the bases of the logs. And:
! --> +
# --> =

Sorry for the unclarity

Gernald · Apr 15, 2008, 07:51 AM

No idea... but if you go to this website: Logarithm it will help you do it.
It's been so long since I've done any, but this way you can ask your teacher later but still get your homework questions right.

ISneezeFunny · Apr 15, 2008, 07:51 AM

It's the Law Too — the Laws of Logarithms

Scroll down and it should give you all the laws of logarithms

galactus · Apr 15, 2008, 01:34 PM

You probably faired well from the link, but here is a way to go about it. Hope it helps, so you can apply it on similar problems.

$log_{4}(2)+log_{9}(2x+5)=log_{8}(64)$

We apply the change of base formula.

Rewrite as:

$\frac{log(2)}{log(4)}+\frac{log(2x+5)}{log(9)} \;\ = \;\ \frac{log(64)}{log(8)}$

Note that 4 and 64 are prefect squares:

$\frac{log(2)}{2log(2)}+\frac{log(2x+5)}{2log(3)} = \;\ \frac{2log(8)}{log(8)}$

$\frac{1}{2}+\frac{log(2x+5)}{2log(3)}=2$

$\frac{log(2x+5)}{log(3)}=3$

$log(2x+5)=3log(3)$

$2x+5=10^{3log(3)}$

Finish?

vinotht · May 24, 2011, 07:44 AM

You can calculate it easily by using online calculator like Log Calculator AntiLog Calculator