# Simplification of Logs

• Dec 9th 2009, 10:51 AM
mikewhant
Simplification of Logs
logbase10(1/10) − logbase10(10/27) − logbase10 1000

I know all the rules for simplifying logs, however I cant seem to get this one right!

If any one could show me some working out towards the end of the simplification that would be great.

Thanks in advance for any help.

Mike
• Dec 9th 2009, 10:54 AM
e^(i*pi)
Quote:

Originally Posted by mikewhant
logbase10(1/10) − logbase10(10/27) − logbase10 1000

I know all the rules for simplifying logs, however I cant seem to get this one right!

If any one could show me some working out towards the end of the simplification that would be great.

Thanks in advance for any help.

Mike

$\displaystyle log_{10}(0.1) = -1$

$\displaystyle log_{10}\left(\frac{10}{27}\right) = log_{10}(10) - 3log_{10}(3)$ (because $\displaystyle 27 = 3^3$)

$\displaystyle log_{10}(1000) = log_{10}(10^3) = 3$

So your sum reduces to $\displaystyle -1-(1-3log_{10}(3))-3$
• Dec 9th 2009, 11:00 AM
mikewhant
Thats great, just one further question, why is it + 3 on the end?

I can see now that log 1000 = 3, however in the original expression, it was
- log 1000, so should'nt it bee -3?

Thaanks again

Mike
• Dec 9th 2009, 11:03 AM
e^(i*pi)
Quote:

Originally Posted by mikewhant
Thats great, just one further question, why is it + 3 on the end?

I can see now that log 1000 = 3, however in the original expression, it was
- log 1000, so should'nt it bee -3?

Thaanks again

Mike

Yes it should, that was an oversight on my part (Thinking)

Thanks for pointing it out :)
• Dec 9th 2009, 11:06 AM
mikewhant
Ah right ok, lol, funny because what was stumping me in the first place was the text book i have the question from gives the same answer you gave, + 3!

Anyway, thanks for that, all cleared up now!

Take care

Mike