Without a calculator evaluate:
Log(base 16)1/8

.... I just used the change of base forumla to get Log1/8 over Log16

is this all that can be done? Thanks for help

2. Originally Posted by vagina69
Without a calculator evaluate:
Log(base 16)1/8

.... I just used the change of base forumla to get Log1/8 over Log16

is this all that can be done? Thanks for help
One of the principles of logs is $\displaystyle \log\frac{x}{y}=\log x-\log y$.

Since the $\displaystyle \log_{x}1=0$ for any $\displaystyle x$, you can simplify to $\displaystyle 0-\log_{16}8$.

There are probably a few ways of going on, but the easiest to me is splitting 8 into $\displaystyle \frac{16}{2}$. The $\displaystyle \log_{16}16=1$ and so we are left with $\displaystyle 0-1-\log_{16}2$.

As 2 is the fourth root of 16, $\displaystyle 16^\frac{1}{4}=2$, and so the $\displaystyle \log_{16}2=\frac{1}{4}$. We can now solve $\displaystyle 0-1-\frac{1}{4}=-\frac{3}{4}$

3. ## logs

Originally Posted by ChristopherDunn
One of the principles of logs is $\displaystyle \log\frac{x}{y}=\log x-\log y$.

Since the $\displaystyle \log_{x}1=0$ for any $\displaystyle x$, you can simplify to $\displaystyle 0-\log_{16}8$.

There are probably a few ways of going on, but the easiest to me is splitting 8 into $\displaystyle \frac{16}{2}$. The $\displaystyle \log_{16}16=1$ and so we are left with $\displaystyle 0-1-\log_{16}2$.

As 2 is the fourth root of 16, $\displaystyle 16^\frac{1}{4}=2$, and so the $\displaystyle \log_{16}2=\frac{1}{4}$. We can now solve $\displaystyle 0-1-\frac{1}{4}=-\frac{3}{4}$
Hi Cris,

Answer could be clearer and more direct.

log base 16 (1/8 ) =logbase 16 ( 2/16)= log base 16 (2) -logbase16 (16)

= 1/4 - 1

bjh

4. Originally Posted by bjhopper
Hi Cris,

Answer could be clearer and more direct.

log base 16 (1/8 ) =logbase 16 ( 2/16)= log base 16 (2) -logbase16 (16)

= 1/4 - 1

bjh

Ah. Right you are. I did add an extra step in there.