sysD Posted January 29, 2011 Share Posted January 29, 2011 Heyo, Anyone aware of a calculator out there on the internet with capacity to calculate logarithms with non-common bases? i.e. log(base2)4(16^(1/3)) - log(base4)0.25^(1/3) = x Yes, yes, I know x=11/3 I just want a quick way to check my answers. Alternatively, is there a quick and easy way to convert non-common bases to common bases? (base10) Link to comment Share on other sites More sharing options...
the tree Posted January 29, 2011 Share Posted January 29, 2011 (edited) Anyone aware of a calculator out there on the internet...No matter what the question here is, the answer is Wolfram Alpha.Alternatively, is there a...Aaand when you just need to look something up, Wikipedia. Edited January 29, 2011 by the tree Link to comment Share on other sites More sharing options...
timo Posted January 29, 2011 Share Posted January 29, 2011 Alternatively, is there a quick and easy way to convert non-common bases to common bases? (base10) Wikipedia sais [math] \log_b r = \frac{\log_a r}{\log_a b} [/math] Link to comment Share on other sites More sharing options...
sysD Posted January 31, 2011 Author Share Posted January 31, 2011 Thx giaz the answer is Wolfram Alpha. oh, and i love you. Link to comment Share on other sites More sharing options...
sysD Posted February 2, 2011 Author Share Posted February 2, 2011 Okay, erm, having some trouble. Here's the question: An investment pays 6%/annum Principle=$3000 Final=$6000 Time= (?) Here's what I keep getting, but something seems very, very wrong: F = P (1+i)^t $6000 = $3000 (1.06)^t $2000 = 1.06^t one way... t = log_1.06(2000) t = (log2000)/(log1.06) t = 130.445 another way... log2000 = log1.06^t log2000 = (t) log1.06 log2000/log1.06 = 130.445 What am I doing wrong here? There's no way it takes 130 years, 5 months, ~10 days, 4 hours, and 48 minutes to double an investment at 6%/annum. Link to comment Share on other sites More sharing options...
timo Posted February 2, 2011 Share Posted February 2, 2011 You're incorrect in assuming that $6000/$3000 = $2000, for instance. Link to comment Share on other sites More sharing options...
sysD Posted February 2, 2011 Author Share Posted February 2, 2011 HAHAH wow. Um. **Fade Out*** Link to comment Share on other sites More sharing options...
sysD Posted February 2, 2011 Author Share Posted February 2, 2011 Bah, I seem to have developed a mental block around logs... can someone tell me if this is right? M=magnitude I=Intensity U=Baseline (M = log (I/U) ) An earthquake registers 6.1 on the Richter Scale. What is the rating on a quake twice as powerful? My answer is : Let "M1" be the quake with a magnitude of 6.1. Let "M2" be the unknown quake magnitude. Let the intensity of M2 be called "I1" Let the intensity of the unknown quake be called "I2" (M1 = log (I/U) ) M1 = log (I/U) I/U = 10^(M1) I = ( 10^(M1) ) * (U) (I2/I1) = 2 The ratio of these two values must = 2 (for a quake twice as intense). (I2/I1) = (( 10^M2 )*( U )) / (( 10^M1 )*( U )) The baseline values ("U") divide out. (I2/I1) = (10^M2 ) / ( 10^M1 ) Substitute value for "M1" (I2/I1) = (10^M2 ) / ( 10^(6.1) ) (I2/I1) = 10^(M2 - 6.1) Substitute value for (I2/I1) 2 = 10^(M2 - 6.1) AKA (in common log form): M2 - 6.1 = log2 Calculator... ( M2 - 6.1 ) = 0.301029996 M2 = ( 0.301029996 + 6.1 ) M2 = 6.401029996 ~ to sig. digits M2 = 6.4 In retrospect I should've just found: 10^(x)=2 where: x=(M2-M1) I seem to be having trouble with logs. Does anyone have some tips to avoid over complicating these problems? Link to comment Share on other sites More sharing options...
timo Posted February 2, 2011 Share Posted February 2, 2011 I don't really understand the question, because I could only guess which variables correspond to "rating" and "power". Anyways, perhaps the identity "log(2x) = log 2 + log x" helps you? Another related and helpful one is [math] \log a^n = n \, \log a [/math], btw. Should both work in all bases. Link to comment Share on other sites More sharing options...
sysD Posted February 2, 2011 Author Share Posted February 2, 2011 well, M is the magnitude, the richter value the question was non-specific... so yeah. i assume its referring to intensity because the richter scale is an exponential function Link to comment Share on other sites More sharing options...
troymius Posted February 12, 2011 Share Posted February 12, 2011 Hi sysD. Yes Wolfram Alfa is amazing. I wrote a little on-line calculator myself. Nothing fancy but does what I need better than any other I could find. It happens to be able to solve your logarithm problem: pascalc.com -you should see the solution by clicking on the link. Link to comment Share on other sites More sharing options...
alpha2cen Posted February 12, 2011 Share Posted February 12, 2011 (edited) In the calculator "log" calculation is carried out by using power series. The problem occurs very small long digit calculation. In this case if you need more accurate value, you had better use more modified method. Edited February 12, 2011 by alpha2cen Link to comment Share on other sites More sharing options...
sysD Posted February 14, 2011 Author Share Posted February 14, 2011 Hi sysD. Yes Wolfram Alfa is amazing. I wrote a little on-line calculator myself. Nothing fancy but does what I need better than any other I could find. It happens to be able to solve your logarithm problem: pascalc.com -you should see the solution by clicking on the link. Haha, thanks man. This is excellent. *bookmarked* Link to comment Share on other sites More sharing options...
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