How to Convert the Base of an Exponent with Logarithms


Home / How to Convert the Base of an Exponent with Logarithms

Introduction to the Logarithm in Mathematics

Logarithm in base b of b to the power x : Image by Mike DeHaan

The logarithm is the inverse operation of exponentiation.

Let’s define “log[b](b^x) = x” for non-zero real numbers ‘b’ and ‘x’. The image here shows how this is usually printed.

We retain the term “base” for logarithms, just as it was for exponents.

In the coin-toss experiment, 2^10 = 1024; so log[2](1024) = 10.

One trivial result is that “log[b](b) = 1” for all non-zero values of ‘b’, since “b = b^1”.

‘e’ is Euler’s Number and Napier’s Constant for Natural Logarithms

Natural Logarithm as a Definite Integral : Image by Mike DeHaan

Say hello to our little friend ‘e’, a favourite of the famous mathematicians Euler and Napier.

The constant ‘e’ is approximately 2.71828…; it is a non-rational number that is the base for natural logarithms.

Mathematicians find ‘e’ and natural logarithms so useful that they reserve the notation “ln(x) = log[e](x)”.

As the image shows, “ln(e)” is defined as the “definite integral from 1 to ‘e’ of dx/x”. As well, since “ln(e) = log[e](e)”, 1 must be the value of that definite integral.

The Reason to use a Logarithm to Convert an Exponent’s Base

At the outset, my problem was to solve for ‘x’ in the equation “2^m = 10^x”, where ‘m’ is a known positive integer and ‘x’ is a positive real.

Thanks to logarithms, this is equivalent to solving “log[2](2^m) = log[10](10^x)”.

The General Rule for Converting Logarithms

Logarithm in base b converts to a Natural Logarithm : Image by Mike DeHaan

The general rule for converting a logarithm from base ‘b’ to natural logarithms in base ‘e’ is “log[b](y) = ln(y)/ln(b)”.

We will use this rule in a later section.

The General Rule for the Logarithm of an Exponential Term

In general, “log[b](x^m) = m*log[b](x).

Therefore “ln(2^m) = m*ln(2)”, which we will need in the next section.

Deriving the Exponents’ Conversion from Base-2 to Base-10

Let’s step through the process for my original problem.

  1. “2^m = 10^x”, to solve for ‘x’.
  2. “m = log[2](2^m)” on the left side.
  3. “x = log[10](10^x)” on the right side.
  4. Apply “log[b](y) = ln(y)/ln(b)” to each side.
  5. “log[2](2^m) = ln(2^m)/ln(2)” on the left.
  6. “log[10](10^x) = ln(10^x)/ln(10)” on the right.
  7. Obviously “log[10](10^x) = x”, to be substituted soon…
  8. But “2^m = 10^x” in /#1/, so we substitute “2^m” for “10^x” in /#6/ as follows…
  9. “x = ln(2^m)/ln(10)” by lines /7/ and /8/.
  10. “x = m*ln(2)/ln(10)”, by the rule for the logarithm of an exponential.

A bit of work on a calculator gives the conversion factor, “ln(2)/ln(10) = 0.69314718055994530941723212145818 / 2.3025850929940456840179914546844 = 0.30102999566398119521373889472446” to a large degree of accuracy.

I’m likely to use 0.301 as the conversion factor.

Let’s check my original problem, which was “2^10 = 1024 = 10^x”. My conversion is “x = 10 * 0.301 = 3.01”.

On the calculator, “10^3.01 = 1023.2929922807541309662751748199″, which is fairly close to 1024.

Leave a Comment