My Math Forum logarithms again!

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 8th, 2010, 10:47 AM #1 Senior Member   Joined: Jul 2009 Posts: 136 Thanks: 0 logarithms again! The textbook answer for log 27 - log 3 is "3." But presuming that the base number is 10 how does this figure? One responder here told me to see it as log 27/3 = log 9, but how does log 9 = 3? I think I'm missing something here.... If it were log(3) 9 then I would understand it better.... But the problem is just log 27 - log 3.... If anyone can explain this, then let me know.... Though I suppose it's possible the text is simply wrong?
 July 8th, 2010, 10:52 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: logarithms again! If you use a base of $\sqrt[3]{9}$ then log 9 = 3.
July 8th, 2010, 11:01 AM   #3
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Re: logarithms again!

Quote:
 Originally Posted by empiricus The textbook answer for log 27 - log 3 is "3." But presuming that the base number is 10 how does this figure? One responder here told me to see it as log 27/3 = log 9, but how does log 9 = 3? I think I'm missing something here.... If it were log(3) 9 then I would understand it better.... But the problem is just log 27 - log 3.... If anyone can explain this, then let me know.... Though I suppose it's possible the text is simply wrong?
Hi empiricus,

I followed up http://www.mymathforum.com/viewtopic.php?f=13&t=14865 in your original post.

July 8th, 2010, 03:47 PM   #4
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Re: logarithms again!

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 Originally Posted by greg1313 If you use a base of $\sqrt[3]{9}$ then log 9 = 3.

OK, but that's the problem--the text question is simply "Log 27 - Log 3." I thought we assume a base of 10 if it is not noted.

 July 8th, 2010, 03:53 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: logarithms again! Like you say, the text could be wrong, or perhaps you are expected to calculate the base given the result of log 9 = 3. I believe it is standard to assume a base of 10 if no base is given, though.
July 8th, 2010, 03:57 PM   #6
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Re: logarithms again!

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 Originally Posted by greg1313 Like you say, the text could be wrong, or perhaps you are expected to calculate the base given the result of log 9 = 3. I believe it is standard to assume a base of 10 if no base is given, though.
Well, that's just it. The answers are in the back and if it were something as Log(b) 9 = 3 then I would say sure, the b = 3, but all it presents is "Log 27 - Log 3" and an answer in the back of "3." Yikes!

 July 8th, 2010, 10:18 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Maybe the intention was log(27)/log(3).
July 9th, 2010, 08:29 AM   #8
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Quote:
 Originally Posted by skipjack Maybe the intention was log(27)/log(3).
Oh, okay. Thanks. But then what would that be? Log (9)? The text answer is "3." I'm assuming at this point it's incorrect.

 July 9th, 2010, 08:31 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: logarithms again! log(27)/log(3) = log(3^3)/log(3) = 3log(3)/log(3) = 3.

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