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 July 7th, 2010, 08:36 AM #1 Senior Member   Joined: Jul 2009 Posts: 136 Thanks: 0 logarithms A question in the text offers the following to simplify without using a calculator: log 27 - log 3. I'm assuming (correctly?) that this would be [10 to the x power = 27] - [10 the x power = 3]. Anyway, the text answer is "3." But this doesn't make sense to me as obviously (once more!) I'm missing something. Any ideas? Thanks.
July 7th, 2010, 09:02 AM   #2
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Re: logartihms

Quote:
 Originally Posted by empiricus A question in the text offers the following to simplify without using a calculator: log 27 - log 3. I'm assuming (correctly?) that this would be [10 to the x power = 27] - [10 the x power = 3]. Anyway, the text answer is "3." But this doesn't make sense to me as obviously (once more!) I'm missing something. Any ideas? Thanks.
Hi empiricus,

Use $\log_b x - \log_b y=\log_b\left(\frac{x}{y}\right)$

$\log (27)-\log (3)=\log\left( \frac{27}{3}\right)=\log (9)$

 July 7th, 2010, 02:33 PM #3 Senior Member   Joined: Jul 2009 Posts: 136 Thanks: 0 Re: logarithms Thanks masters. But how would log9 = 3? That is the problem for me. If you can explain that, this would be great! Thanks.
July 8th, 2010, 10:57 AM   #4
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Re: logarithms

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 Originally Posted by empiricus Thanks masters. But how would log9 = 3? That is the problem for me. If you can explain that, this would be great! Thanks.
Hi empiricus,

You text answer would seem to be incorrect if the log base is 10.

$\log (9)\: \neq \: 3$

$\log(9)= \log (27)\:-\: \log (3)\:\approx\: .9542425094$

If $\log_b (9)=3$, then

$b^3=9$, and

$b=\sqrt[3]{9}$. Then the following woud be correct:

$\log_{\sqrt[3]{9}}(9)=3$

only if the log base was $\sqrt[3]{9}$

 July 8th, 2010, 03:49 PM #5 Senior Member   Joined: Jul 2009 Posts: 136 Thanks: 0 Re: logarithms OK, that was strange. Thanks for showing me that I'm not going crazy! And the one after it reads like this: (log 16 - log 2 ) divided by log 2. The answer for this, according to the text, is also "3." But I am not getting that.... Help!
 July 8th, 2010, 03:59 PM #6 Senior Member   Joined: Jul 2009 Posts: 136 Thanks: 0 Re: logarithms By the way, what if we figure Log 27 - Log 3? If the base, (b) was 3, then wouldn't we figure 3 - 1? But, again, I'm likely doing something wrong....
July 8th, 2010, 04:11 PM   #7
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Re: logarithms

Quote:
 Originally Posted by empiricus (log 16 - log 2 ) divided by log 2. The answer for this, according to the text, is also "3."
What number, when raised to the 3rd power, equals 8?

July 9th, 2010, 08:30 AM   #8
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Re: logarithms

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by empiricus (log 16 - log 2 ) divided by log 2. The answer for this, according to the text, is also "3."
What number, when raised to the 3rd power, equals 8?
"2," but that's not the text answer which is "3." Unless you are trying to show me something else here? Not sure....

 July 9th, 2010, 08:33 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond Re: logarithms (log(16) - log(2))/log(2) = log(/log(2) = log(2^3)/log(2) = 3log(2)/log(2) = 3.
July 17th, 2010, 04:19 AM   #10
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Re: logarithms

Thanks for the clear response but I have some questions as none of this is going by the rules given in the texts.

1. How did you go from log 16 - log 2 to log 8?

2. If one says that it's a matter of log 16/log 2 then what's the point of re-stating it as a subtraction?

3. I would assume that log 16 - log 2 = log 14. I know it's supposed to be the log of 16 - the log of 2, but we do not know that the base number is but 10 here as nothing else is given as the base number.

Quote:
 Originally Posted by greg1313 (log(16) - log(2))/log(2) = log(/log(2) = log(2^3)/log(2) = 3log(2)/log(2) = 3.

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