Posted by Sam on Sunday, October 14, 2012 at 3:14pm.
Using the approximation log10^2=0.301, find:
- Logarithms - Steve, Sunday, October 14, 2012 at 3:27pm
since 10^8 = 10^8,
log 10^8 = 8
giving the exponent makes finding the log easy, since a log is in fact an exponent
log 100 = 2
because 10^2 = 100
- Wrong, typo - Damon, Sunday, October 14, 2012 at 3:29pm
log 10^2 = IS NOT .301
log base 10 of 2 = .301
usually written log10(2)
I bet you want log base 10 of 8
log10 (2^3) = 3 log10(2) = 3(.301) = .903
which i bet is what your are looking for
- Logarithms - Sam, Sunday, October 14, 2012 at 3:36pm
Yes Damon that's what I meant. Thanks!
- Logarithms - my bad - Steve, Sunday, October 14, 2012 at 3:37pm
My bad. I didn't read the entire problem. I might even have caught the strange notation. Go with Damon.
- Logarithms - Sam, Sunday, October 14, 2012 at 3:39pm
It's fine. Thanks
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