1. ## Simplifying log expressions

Hi,
I am having trouble with this one problem. If someone could help I would be extremely grateful...

Use the properties of logs to simplify the following log expression :

log9 (1/8)

My math book says the answer is -1 -log9 2, but I just can not figure it out.

Thanks

2. Originally Posted by scutt
Hi,
I am having trouble with this one problem. If someone could help I would be extremely grateful...

Use the properties of logs to simplify the following log expression :

log9 (1/8)

My math book says the answer is -1 -log9 2, but I just can not figure it out.

Thanks
recheck that "book" answer ... it's not even close.

$\log_9\left(\frac{1}{8}\right) = -\log_9(8)$

you sure that the base of the log is 9 ?

3. $\log_9\left(\frac{1}{8}\right)$

$-1-log_92$

$log_9\frac{1}{8}=-\textrm{log}_98\,\,\,\because\,\,\,log_ab^n=nlog_a b$
$-log_98=-log_9(2^3)\,\,\because\,\,\,2^3=8$
$-log_9(2^3)=-3log_92\,\,\,\because\,\,\,log_ab^n=nlog_ab$
$-3log_92=\frac{-3log_32}{2}\,\,\,\because\,\,\, log_{b^n}a=\frac{log_ba}{n}$

$-1-log_92=-\frac{log_32}{2}-1$
$-\frac{log_32}{2}-1\neq\frac{-3log_32}{2}$

I would say your book is wrong.

You can most likely use anything past the first step and get credit for simplification. :P

My idea of simplification is the lowest possible base and coefficient numbers.