Logarithms with Unknown BasesDate: 12/12/2001 at 17:47:20 From: Juan F. Rodriguez Subject: Logarithms with unknown bases I read the following problem in a textbook: Provided only that, in some bases, log 2 = 0.607 and log 3 = 0.959, find a good value for log 5. Thus far I have solved it using the following identity about logarithms with arbitrary bases: log_a(5)/log_a(3) = ln(5)/ln(3) where log_a stands for the logarithm in base a; then, log_a(5) = ln 5*log_a 3/ln 3 = 1.609*0.959/1.099 = 1.405. However, this solution is unsatisfactory, for two reasons: First, I had to use a table or calculator to find the natural logarithm values. Second, I did not use all the information provided. I am still working on an alternative solution. Thanks for your help. Date: 12/13/2001 at 15:48:50 From: Doctor Peterson Subject: Re: Logarithms with unknown basis Hi, Juan. Interesting problem! But according to the same base change formula you used (using 2 and 3 rather than 3 and 5), if their log(3) is correct, log(2) should be 0.605. Something isn't quite right. If you calculate what base they are using, it looks as if they may have meant to use pi. Then log_pi(2) = 0.6055, and log_pi(3) = 0.9597, which is close to their numbers. One way to estimate log(5) is to interpolate. We can easily find log(4) = log(2^2) = 2 log(2) = 2*0.607 = 1.214 log(6) = log(2*3) = log(2) + log(3) = 0.607 + 0.959 = 1.563 Using linear interpolation, log(5) = (log(4) + log(6))/2 = (1.214+1.563)/2 = 2.777/2 = 1.388 (If I used log(2) = 0.605, I would get log(5) = (3log(2)+log(3))/2 = 1.387.) That's not too bad, and may well be the expected answer. In fact, we can work backward to find what that is really the log of: (log(4) + log(6))/2 = log(24)/2 = log(sqrt(24)) = log(4.899) So the same value could have been found instead by noticing that 24 = 2^3*3 is close to 25 = 5^2, and finding the log of its square root. Can we improve on this by finding, say, a larger power of 5 that is close to a number with prime factors 2 and 3 only? Playing around with numbers, I found log(5) = log(2.5) + log(2) ~ log(2.56) + log(2) = log(256) - log(100) + log(2) = 8 log(2) - 2 log(2) - 2 log(5) + log(2) = 7 log(2) - 2 log(5) so log(5) ~ 7/3 log(2) = 1.416 (Again, using log(2) = 0.605, we get log(5) ~ 1.412, just a bit closer to the correct value of 1.405.) This is actually 7/3 log(2) = log(cbrt(2^7)) = log(5.04) so it's clear why this is a better approximation. I have no idea what is considered "a good value," but we could certainly improve this as much as we have time for. In fact, I found on my calculator that 3^35 is close to 5*10^16 (within 3 significant digits), and used that to get a value of 1.405 for log(5), using log(2) = 0.605. Not bad! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/