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 $\log\frac{x}{y}=\log x-\log y$.

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

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

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

3. ## logs

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

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

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

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