logarithms?!

lemon301 · #1 $Old$ January 2nd, 2008, 07:36 AM

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.

earboth · #2 $Old$ January 2nd, 2008, 08:09 AM

Quote:

Originally Posted by lemon301 $View Post$

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.

lemon301 · #3 $Old$ January 2nd, 2008, 10:45 AM

Quote:

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?

earboth · #4 $Old$ January 2nd, 2008, 10:55 PM

Quote:

Originally Posted by lemon301 $View Post$

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

$\log_3(27)=3$

topsquark · #5 $Old$ January 3rd, 2008, 01:32 PM

Quote:

Originally Posted by lemon301 $View Post$

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