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 $100$?" $${\log }_{10}(100)=x\iff 10^x=100$$ Here, $x=2.$ Suppose instead that the value of $$lo{g}_{10}(-1)$$ 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 $$10^3=10\cdot 10\cdot 10.$$ 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. $$x^{\text{-}}n=\frac{1}{x^n}$$ Can $10^x$ be negative if $x$ is negative? A base with a negative exponent can be rewritten as a fraction.

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.