LOGARITHM ARGUMENTS OF LESS THAN OR EQUAL TO ZERO. IS IT POSSIBLE?
Copyright by Ingrid Stewart, Ph.D. Please Send Questions and Comments to ingrid.stewart@csn.edu.
What is the value of logb0?
As we know, a logarithm can only be defined if the base b is greater than 0.
For a moment, let's assume that a logarithm argument equals 0, and let's logb0 be equal to some number n, that is, logb0 = n.
In its exponential form this would be 0 = bn.
You are challenged to find a base b and an exponent n for which this is true!!! What about if n euqals 0? But any number b raised to a 0 power has a value of 1. What about if n is a negative number? But any number b raised to a negative number is a fraction.
You will find that there is no such power n that would make an exponential expression be equal to 0 given a base b greater than 0.
Therefore, logb0 is not defined.
What is the value of logb(a)?
Logarithms of negative numbers are not defined in the real numbers. If you are expected to find the logarithm of a negative number, an answer of "undefined" is sufficient in most cases.
However, it is possible to evaluate a logarithm of negative numbers. The answer will be a complex number, a number of the form a + bi, where i is the imaginary number equal to .
For a moment, let's assume that the logarithm argument is a negative number, say a, and let's logb(a) be equal to some number n, that is, logb(a) = n. We will further assume that base b is not equal to the number e.
Now let's use the Change-of-Base Property to change logb(a) to base e.
Note that loge (a) is the the same thing as loge (1 a). We can now use the Product Rule of logarithms to write the numerator as follows:
loge (a) = loge (1 a)
= loge (1) + loge a
and
Now the only problem is figuring out what loge (1) is equal to. It might look like an impossible thing to evaluate at first, but there is a pretty famous equation known as Euler's Identity that can help us.
Euler's Identity states:
This result comes from Calculus and is a power series expansions of sine and cosine. We won't explain that too in-depth, but if you are interested, there is a nice page here which explains a bit more.
For now, let us simply take the natural logarithm of both sides of Euler's Identity:
Using th basic logarithm property logbbx = x, we find
So now that we know what loge (1) is equal to, we can substitute it back into the equation
That is,
Now we have a formula for finding logarithms of negative numbers.
For example, if we were asked to find log2(10) and round the answer to three decimal places, we would get