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Logarithms with Unknown Bases


Date: 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/   
    
Associated Topics:
High School Logs

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