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This is a really, very simple question, but I've never been an extremely confident mathematician and I just want to make sure that my attempt was correct. Oh and this is homework incase you were thinking I'm trying to sneak answers. :) All logarithms are to base b . The original expression is: $$ \log(4) - 3\log(1/3) + \log(2) $$

So I decided, the first thing to do was to invoke the power law so they're all in the same form:

$$ \log(4) - \log(1/3^3) + \log(2) = \log(4) - \log(1/27) + \log(2) $$

Then I used the subtraction law:

$$ \log(4 \cdot 27) + \log(2) = \log(108) + \log(2) $$

And finally I applied the addition law:

$$ \log(108 \cdot2) = \log(216) $$

The question was to simplify it to a single logarithm. I just wanted to ensure I had done this right. All the other answers in the paper evaluate to like $log(5)$ and 216 seemed a little out of place :).

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If it's homework, you should just add the (homework) tag, rather than mention it in the body of your post. This helps us out quite a bit (for sorting posts, etc.). –  Cameron Buie Mar 13 '13 at 15:24
    
Sorry, I'll ammend it now :) –  notverygoodatmaths Mar 13 '13 at 15:26
1  
To show multiplication, use \cdot to get $4 \cdot 27$. The period looks too much like a decimal point. Also \log will get you $\log$ which is the proper font for functions. –  Ross Millikan Mar 13 '13 at 15:29
    
Okay, thanks. I tried looking for tutorals on LaTex but I couldn't find any so I just did my best :P –  notverygoodatmaths Mar 13 '13 at 15:30
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Thanks for trying the $\LaTeX$. This site is a good place to learn. You can right-click on any you see and select Show Math as -> TeX commands to see how it was done –  Ross Millikan Mar 13 '13 at 15:34

2 Answers 2

up vote 1 down vote accepted

Yes you've done that correctly. I like to pull the negative sign up with the exponent to turn everything into addition, but what you've done is fine:

$$\log(4) - 3\log(1/3) + \log(2) = \log(4) + \log((1/3)^{-3}) + \log(2)$$

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Nicely done. ${}{}{}{}{}{}{}{}$

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So it's correct then? –  notverygoodatmaths Mar 13 '13 at 15:26
    
Yep! ${}{}{}{}$ –  Cameron Buie Mar 13 '13 at 15:28
    
Ah excellent! I've been programming for years, and I've gotten good at that. I've just got gaping holes in my math knowledge. Thanks for that :) –  notverygoodatmaths Mar 13 '13 at 15:29

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