Logarithms, written in the form ${\log }_b(m),$ are only defined for m>0. To prove this, it is necessary to go back to the definition of logarithms. The logarithm of a number is the exponent to which the logarithm's base must be raised in order to get the desired number. For instance, ${\log }_{10}(100)$ is the same as asking the question, "What is the exponent needed to raise 10 to in order to get 100?" Here, x=2. Suppose instead that the value of wanted to be found. Exponentially, this can be though of as $10^x=\text{-}1.$ The graph of $y=10^x$ can be used to visualize this equation.

From the diagram, it can be seen that $10^x$ never lies below the x-axis. This means the function value is never negative. In fact, the function never equals 0 — it gets closer and closer to 0. To examine this, let's consider positive and negative values of x.

$10^x$ is 10 multiplied by itself x times. For example Since 10 is a positive number, no matter how many times it is multiplied, the product will always be positive. Therefore, there are no positive x-values that make $10^x$ negative. Any base raised to a negative exponent can be rewritten as a fraction. Can $10^x$ be negative if x is negative? A base with a negative exponent can be rewritten as a fraction.

x	-1	-2	-3	-4
$10^x$	$10^{\text{-}1}$	$10^{\text{-}2}$	$10^{\text{-}3}$	$10^{\text{-}4}$
Fraction	$\frac{1}{10^1}$	$\frac{1}{10^2}$	$\frac{1}{10^3}$	$\frac{1}{10^4}$
=	0.1	0.01	0.001	0.0001

The "more negative" x becomes the smaller $10^x$ will be. It gets closer and closer to 0 without actually equaling 0 or becoming negative.

Therefore, there are no values of x that make $10^x$ negative. Since $10^x=\text{-}1$ has no real solution, ${\log }_{10}\text{-}1$ is undefined.