# Logs - Without calculator

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• Jan 16th 2009, 03:52 PM
Peleus
Logs - Without calculator
Hi all,

What's the best way to work out log's without a calculator / other aids?

The question is $\displaystyle \log_{8}\frac{1}{4}$ what is the answer.

I know the answer is $\displaystyle -\frac{2}{3}$ and I can work it out simply on a calculator. How do I do it without though?

I'm not sure the working method (without trial and error), any help would be appreciated. Sorry if it's pretty basic.
• Jan 16th 2009, 04:00 PM
Jhevon
Quote:

Originally Posted by Peleus
Hi all,

What's the best way to work out log's without a calculator / other aids?

The question is $\displaystyle \log_{8}\frac{1}{4}$ what is the answer.

I know the answer is $\displaystyle -\frac{2}{3}$ and I can work it out simply on a calculator. How do I do it without though?

I'm not sure the working method (without trial and error), any help would be appreciated. Sorry if it's pretty basic.

use the change of base formula

recall, $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

so $\displaystyle \log_8 \frac 14 = \log_8 (4^{-1}) = - \log_84 = - \frac {\log_24}{\log_28} = - \frac 23$

for the second equal sign, i used the rule: $\displaystyle \log_a (x^n) = n \log_a x$
• Jan 16th 2009, 04:03 PM
Peleus
Awesome, makes it a lot clearer. Thank you.

Sometimes trying to self teach yourself things you run into basic questions you don't know the answer for :)

Appreciate it.
• Jan 16th 2009, 05:25 PM
skeeter
let $\displaystyle y = \log_8\left(\frac{1}{4}\right)$

change to an exponential equation ...

$\displaystyle 8^y = \frac{1}{4}$

$\displaystyle (2^3)^y = 2^{-2}$

$\displaystyle 2^{3y} = 2^{-2}$

$\displaystyle 3y = -2$

$\displaystyle y = -\frac{2}{3}$
• Jan 16th 2009, 06:55 PM
Peleus
I'll put another one in here instead of starting up a new thread.

I need to simplify this logarithm. (By solving)

$\displaystyle \log_{2}18 - 2\log_{2}3$

So far I change that (I don't know if it breaks rules or not) to ...

$\displaystyle -2\log_{2}\frac{18}{3} = -2\log_{2}6$

That gives me a nasty decimal answer which I know isn't right. Correct answer is listed as one but I'm not sure how to manipulate it to cancel it out.

Any ideas?

Thank you again for your help so far.
• Jan 16th 2009, 07:05 PM
Jhevon
Quote:

Originally Posted by Peleus
I'll put another one in here instead of starting up a new thread.

I need to simplify this logarithm. (By solving)

$\displaystyle \log_{2}18 - 2\log_{2}3$

So far I change that (I don't know if it breaks rules or not) to ...

$\displaystyle -2\log_{2}\frac{18}{3} = -2\log_{2}6$

That gives me a nasty decimal answer which I know isn't right. Correct answer is listed as one but I'm not sure how to manipulate it to cancel it out.

Any ideas?

Thank you again for your help so far.

yes, you broke the rules. you have to change $\displaystyle 2 \log_2 3$ to $\displaystyle \log_2 (3^2) = \log_2 9$ BEFORE combining the logs.

look at the rule again for simplifying the difference of two logarithms. note that the constant in front of each is 1

P.S. you should post new questions in a new thread
• Jan 16th 2009, 07:22 PM
Peleus
Excellent.

I know how to handle numbers in front now, thank you =)

I'll try and put new questions in new threads now, didn't want to clog the forums with my basic questions though ;)
• Jan 16th 2009, 07:57 PM
Jhevon
Quote:

Originally Posted by Peleus
Excellent.

I know how to handle numbers in front now, thank you =)

glad to hear it.

Quote:

I'll try and put new questions in new threads now, didn't want to clog the forums with my basic questions though ;)
don't worry about "clogging" the forum. the mods are here to make sure that does not happen. in fact, putting new questions in new threads keep the forum organized and so helps with keeping the forum clog free. in addition, you are more likely to get your question answered quickly if you post in a new thread. (users may ignore the thread if they see that someone has responded to it, especially if that someone is a Helper. so if that helper is not around, you won't get your questions answered as quickly as possible)

take care. good luck with your studies