Since every exponential function $f(x) = b^x$, where $b \gt 0, b \neq 1$, passes the horizontal line test, each such function must be invertible.

Let us refer to the inverse of $f$ as the **logarithm base $b$ of $x$**. Further, let us denote this function $f$ by $\log_b x$, leaving off the parentheses around the input $x$ unless they are needed. Notice, this mimics the style in which we write trigonometric functions.

Of course, by virtue of being $f$'s inverse, we know that

$$y=\log_b x \quad \textrm{if and only if} \quad b^y=x$$In the special case when $b=e$, we write $y=\ln x$, saying that $y$ is the **natural logarithm** of $x$. This then likewise requires

Recalling that the graph of an inverse function is found by reflecting the original function over the line $y=x$, we discover that the graphical nature of a logarithmic function also depends largely on whether the base $b$ is greater than or less than one, as the graphs below suggest.

Several properties of log functions are immediately evident from both the above graphs, such as:

The domain consists of only positive real values, $(0,\infty)$

The range is the set of all reals, $\mathbb{R}$

The only intercept is $(1,0)$

The line $x=0$ (i.e., the $y$-axis) is a vertical asymptote

The function increases on its entire domain if $b \gt 1$, and decreases on the same if $0 \lt b \lt 1$

The function is one-to-one

Other properties follow immediately from the inverse relationship with a related exponential function:

$$\begin{array}{rcll} \log_b b &=& 1 & \scriptsize{\textrm{since } b^1 = b}\\ \log_b 1 &=& 0 & \scriptsize{\textrm{since } b^0 = 1}\\ \log_b b^x &=& x &\scriptsize{\textrm{since for inverse functions } f \textrm{ and } f^{-1} \textrm{ we know that } f^{-1}(f(x)) = x}\\ b^{\log_b x} &=& x & \scriptsize{\textrm{since for inverse functions } f \textrm{ and } f^{-1} \textrm{ we know that } f(f^{-1}(x)) = x} \end{array}$$Of course, in the special case where the base is $e$, the above four properties take the following forms:

$$\begin{array}{rcl} \ln e &=& 1\\ \ln 1 &=& 0\\ \ln e^x &=& x\\ e^{\ln x} &=& x \end{array}$$Indeed, the inverse relationship between logarithmic and exponential functions allows us to write any equation in the exponential form $b^y=x$ in an equivalent logarithmic form, and vice-versa, as shown below:

$$\begin{array}{rclcrcl} \log_3 9 &=& 2 &\longleftrightarrow& 3^2 &=& 9\\ \log_{10} 0.00001 &=& -5 &\longleftrightarrow& 10^{-5} &=& 0.00001\\ \log_8 4 &=& \frac{2}{3} &\longleftrightarrow& 8^{2/3} &=& 4\\ \log_{1/2} 4 &=& -2 &\longleftrightarrow& \left( \frac{1}{2} \right)^{-2} &=& 4\\ \log_b \sqrt{b} &=& \frac{1}{2} &\longleftrightarrow& b^{1/2} &=& \sqrt{b}\\ \ln 1 &=& 0 &\longleftrightarrow& e^0 &=& 1\\ \log_3 c &=& N &\longleftrightarrow& 3^N &=& c\\ \ln 4 &=& x &\longleftrightarrow& e^x &=& 4 \end{array}$$As the above demonstrate you can think of $\log_b x$ as "the exponent that must be placed on $b$ to produce $x$". Remembering this phrase can often make evaluating logarithms quick and easy to do in one's head.

Taking this idea a step further, if logarithms are exponents as the phrase above suggests, then the laws of exponents will some force analogous laws for logarithms.

Three such laws/properties for logarithms are given below. (Click the links provided to see proofs of each.)

$\displaystyle{\log_b xy = \log_b x + \log_b y}$

First note that we can evaluate $\log_b x$ and $\log_b y$, since $x,y,$ and $b$ are positive real values with $b \neq 1$. Let us denote these two values by $m$ and $n$, so that

$$\log_b x = m \quad \textrm{and} \quad \log_b y = n$$If we express the above in exponential form, we have

$$b^m = x \quad \textrm{and} \quad b^n = y$$With the above, the rest of the proof follows quickly, as shown below. Notice how the proof hinges on the fact that $b^m b^n = b^{m+n}$, one of our laws of exponents.

$$\begin{array}{rcl} \log_b xy &=& \log_b b^m b^n \\ &=& \log_b b^{m+n}\\ &=& m + n\\ &=& \log_b x + \log_b y \end{array}$$$\displaystyle{\log_b \frac{x}{y} = \log_b x - \log_b y}$

First note that we can evaluate $\log_b x$ and $\log_b y$, since $x,y,$ and $b$ are positive real values with $b \neq 1$. Let us denote these two values by $m$ and $n$, so that

$$\log_b x = m \quad \textrm{and} \quad \log_b y = n$$If we express the above in exponential form, we have

$$b^m = x \quad \textrm{and} \quad b^n = y$$With the above, the rest of the proof follows quickly, as shown below. Notice how the proof hinges on the fact that $\displaystyle{\frac{b^m}{b^n} = b^{m-n}}$, one of our laws of exponents.

$$\begin{array}{rcl} \log_b \frac{x}{y} &=& \log_b \frac{b^m}{b^n}\\ &=& \log_b b^{m-n}\\ &=& m - n\\ &=& \log_b x - \log_b y \end{array}$$$\displaystyle{\log_b x^n = n \log_b x}$

First note that we can evaluate $\log_b x$, since $x$ and $b$ are positive real values with $b \neq 1$. Let us denote this value by $m$, so

$$\log_b x = m$$Expressing the above in exponential form, we then have $$b^m = x$$

Now the rest of the proof follows quickly, as shown below. Notice how the proof hinges on the fact that $(b^m)^n = b^{mn}$, one of our laws of exponents.

$$\begin{array}{rcl} \log_b x^n &=& \log_b (b^m)^n \\ &=& \log_b b^{mn}\\ &=& mn\\ &=& n\log_b x \end{array}$$

Another very useful property of logarithms derived from the laws of exponents is the **Change of Base Theorem**, shown below. This theorem allows one to rewrite any logarthm (say in base $y$) in terms of logarithms involving any desired legal base $b$ that one wishes.

First note that we can evaluate $\log_b x, \log_b y$ and $\log_y x$, since $x,y,$ and $b$ are all positive real values with $b,y \neq 1$. Let us denote the last two values by $m$ and $n$, so that

$$\begin{array}{rcl} \log_b y &=& m\\ \log_y x &=& n\\ \end{array}$$If we express the above in exponential form, we have

$$\begin{array}{rcl} b^m &=& y\\ y^n &=& x\\ \end{array}$$Note, that we know from the laws of exponents that

$$b^{mn} = (b^m)^n = y^n = x$$Rewriting the resulting $b^{mn} = x$ in logarithmic form, we have

$$\log_b x = mn$$Now, we can substitute for $m$ and $n$ to obtain

$$\log_b x = (\log_b y)(\log_y x)$$Finally, dividing both sides by $\log_b y$ completes the proof.

$$\frac{\log_b x}{\log_b y} = \log_y x$$