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Old 01-02-2008, 07:36 AM
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Default logarithms?!

I never caught on to logarithms, so anything would help. I need help solving these two problems, but if anyone can direct me somewhere that really explains logarithms it would be a great help!

Find the value of log 50 - log 5, where log is the base 10 logarithm.

Evaluate log(v3)27.
I didnt really know how to type that out. It should be log (down 3) 27. Would the down 3 be the base? I just don't understand what to do with it.
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Old 01-02-2008, 08:09 AM
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I never caught on to logarithms, so anything would help. I need help solving these two problems, but if anyone can direct me somewhere that really explains logarithms it would be a great help!

Find the value of log 50 - log 5, where log is the base 10 logarithm.

Evaluate log(v3)27.
I didnt really know how to type that out. It should be log (down 3) 27. Would the down 3 be the base? I just don't understand what to do with it.
Hello,

\log(a) - \log(b) = \log\left(\frac ab\right)

Use this property to do your problem. You should come out with 1.

\log_{3}(27) = \log_{3}(3^3) Now you can easily determine the value of the logarithm.
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Old 01-02-2008, 10:45 AM
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Originally Posted by earboth View Post
Hello,

\log(a) - \log(b) = \log\left(\frac ab\right)

Use this property to do your problem. You should come out with 1.

\log_{3}(27) = \log_{3}(3^3) Now you can easily determine the value of the logarithm.
So \log_{3}(3^3) would just simplify out to log 3, correct?
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Old 01-02-2008, 10:55 PM
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So \log_{3}(3^3) would just simplify out to log 3, correct?
Hello,

logarithm is only another word for exponent BUT you have to add the base to which this exponent belongs:
Example: If you say the exponent is 3 then you can get 1000 = 10^3 or \frac18 = \left(\frac12\right)^3 . In both cases the exponent is 3.

\log_3(27) aks for an exponent which belongs to the base 3 so that the result of the power is 27.

Comparing my previous post you'll see that you have to use the exponent 3 with the base 3 to get 27. Therefore

\log_3(27)=3
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Old 01-03-2008, 01:32 PM
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So \log_{3}(3^3) would just simplify out to log 3, correct?
If you like, there is another rule you can use:
log_a(b^n) = n \cdot log_a(b)

So
log_3(3^3) = 3 \cdot log_3(3) = 3 \cdot 1 = 3

-Dan
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