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### Logarithms with Unknown Bases

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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.
```

```
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

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Logs

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