# Solving Logarithms

• Oct 31st 2007, 05:35 PM
cjmac87
Solving Logarithms
How would I go about solving for "x" for something like this (should be fairly simple):

$\displaystyle log7^x=log7^9$

edit: thats supposed to be base 7. how would you enter the code for that?
• Oct 31st 2007, 05:38 PM
Jhevon
Quote:

Originally Posted by cjmac87
How would I go about solving for "x" for something like this (should be fairly simple):

$\displaystyle log7^x=log7^9$

if log(something) = log(something) then the somethings must be equal, the log function is one-to-one.

so just drop the logs. you have 7^x = 7^9, now what is x?
• Oct 31st 2007, 06:01 PM
cjmac87
thanks that helps.

ok what about something thats not one-to-one:
$\displaystyle log_2{2^-1}=x$
• Oct 31st 2007, 06:17 PM
Jhevon
Quote:

Originally Posted by cjmac87
thanks that helps.

ok what about something thats not one-to-one:
$\displaystyle log_2{2^-1}=x$

you should know that $\displaystyle \log_aa^n = n$
• Oct 31st 2007, 06:55 PM
cjmac87
ah, oh course. very simple.

one last thing i need cleared up which is probably easy to:

$\displaystyle log_2{16}$ can be solved how?
how about $\displaystyle log_{16}{1/4}$?
• Oct 31st 2007, 07:00 PM
kalagota
Quote:

Originally Posted by cjmac87
ah, oh course. very simple.

one last thing i need cleared up which is probably easy to:

$\displaystyle log_2{16}$ can be solved how?
how about $\displaystyle log_{16}{1/4}$?

$\displaystyle let \, \log_2 {16} = x$

then

$\displaystyle 2^x = 16 = 2^4$
now, you can continue..

the other one, note that

$\displaystyle \log_a {\frac{c}{d}} = \log_a {c} - \log_a {d}$

and $\displaystyle \, \log_a 1 = 0 \, for \, all \, a>0$

can you do the rest?
• Oct 31st 2007, 07:03 PM
TKHunny
You seem to be missing an important concept:

Logarithm = Exponent

Trust me on this. What 2 roles are played by the 4, below?

$\displaystyle log_{2}(16)\;=\;4\;\iff\;2^{4}\;=\;16$

You do the other one.