Logarithmic Functions

A logarithm undoes an exponent, so it’s no surprise that the graph of $y = \log_b x$ is the graph of $y = b^x$ “undone”: its reflection in the line $y = x$. Once you see that, every key feature of a logarithmic graph comes straight from a feature of the exponential function you already know.

Key ideas

The inverse of an exponential function

To find the inverse of $y = b^x$, swap $x$ and $y$, then solve for $y$:

$$x = b^y \quad\Longleftrightarrow\quad y = \log_b x$$

So $y = \log_b x$ is the inverse of $y = b^x$. As with any inverse, the points swap coordinates: $(2, 4)$ on $y = 2^x$ becomes $(4, 2)$ on $y = \log_2 x$.

$y = 2^x$	$\left(-2, \tfrac{1}{4}\right)$	$\left(-1, \tfrac{1}{2}\right)$	$(0, 1)$	$(1, 2)$	$(2, 4)$	$(3, 8)$
$y = \log_2 x$	$\left(\tfrac{1}{4}, -2\right)$	$\left(\tfrac{1}{2}, -1\right)$	$(1, 0)$	$(2, 1)$	$(4, 2)$	$(8, 3)$

$y = \log_2 x$ is the reflection of $y = 2^x$ in the line $y = x$.

Why the inverse is a function

An exponential function is always increasing (if $b \gt 1$) or always decreasing (if $0 \lt b \lt 1$), so it passes the horizontal line test: no output happens twice. That means its inverse passes the vertical line test, so the inverse is a function. Compare $y = x^2$, which gives the output $4$ for both $x = 2$ and $x = -2$, so its inverse is not a function.

Key features of y = log_b x

Every feature of the logarithmic graph is a swapped feature of the exponential graph.

Feature	$y = b^x$	$y = \log_b x$
Domain	$\{x \in \mathbb{R}\}$	$\{x \in \mathbb{R} \mid x \gt 0\}$
Range	$\{y \in \mathbb{R} \mid y \gt 0\}$	$\{y \in \mathbb{R}\}$
Intercept	$y$-intercept $1$	$x$-intercept $1$
Asymptote	horizontal, $y = 0$	vertical, $x = 0$
Key points	$(0, 1)$ and $(1, b)$	$(1, 0)$ and $(b, 1)$

A few things to notice about $y = \log_b x$:

• There’s no $y$-intercept and no horizontal asymptote.
• The graph hugs the $y$-axis as $x$ gets close to $0$, but never touches it.
• It keeps rising forever, but very slowly: $\log x$ only reaches $6$ when $x = 1\,000\,000$.

Increasing or decreasing

• If $b \gt 1$, $y = \log_b x$ is increasing: negative for $0 \lt x \lt 1$, positive for $x \gt 1$.
• If $0 \lt b \lt 1$, $y = \log_b x$ is decreasing: positive for $0 \lt x \lt 1$, negative for $x \gt 1$.

Either way, the graph passes through $(1, 0)$ and $(b, 1)$ and has the vertical asymptote $x = 0$.

Comparing bases

For $x \gt 1$, a bigger base gives a flatter graph. Base $\tfrac{1}{2}$ is the mirror image of base $2$ in the $x$-axis.

For $x \gt 1$, a larger base gives a smaller log, so the graph is flatter: $\log_2 8 = 3$, but $\log 8 \approx 0.90$. That makes sense: a bigger base needs a smaller exponent to reach the same number.

Worked examples

Example 1: Graphing from the exponential

Graph $y = \log_3 x$ using a table of values for $y = 3^x$. State the domain, range, intercept, and asymptote.

Solution. Make a table for $y = 3^x$, then swap the coordinates.

$y = 3^x$	$\left(-2, \tfrac{1}{9}\right)$	$\left(-1, \tfrac{1}{3}\right)$	$(0, 1)$	$(1, 3)$	$(2, 9)$
$y = \log_3 x$	$\left(\tfrac{1}{9}, -2\right)$	$\left(\tfrac{1}{3}, -1\right)$	$(1, 0)$	$(3, 1)$	$(9, 2)$

Plot the second row and join the points with a smooth curve that rises slowly to the right and drops steeply toward the $y$-axis as $x$ gets close to $0$.

• Domain: $\{x \in \mathbb{R} \mid x \gt 0\}$
• Range: $\{y \in \mathbb{R}\}$
• $x$-intercept: $1$; no $y$-intercept
• Vertical asymptote: $x = 0$

Example 2: Finding inverses

Find the inverse of each function, and state the domain and range of the inverse.

• (a) $f(x) = 6^x$
• (b) $g(x) = \log_{0.5} x$

Solution.

(a) Write $x = 6^y$ and switch to log form: $f^{-1}(x) = \log_6 x$. The domain of $f^{-1}$ is the range of $f$: $\{x \in \mathbb{R} \mid x \gt 0\}$. Its range is the domain of $f$: $\{y \in \mathbb{R}\}$.

(b) Write $x = \log_{0.5} y$ and switch to exponential form: $g^{-1}(x) = 0.5^x$. The domain of $g^{-1}$ is $\{x \in \mathbb{R}\}$ and its range is $\{y \in \mathbb{R} \mid y \gt 0\}$.

Check (a) with a number: $f(2) = 36$, and $f^{-1}(36) = \log_6 36 = 2$. ✓

Example 3: Base 2 versus base one half

Show that $\log_{\frac{1}{2}} x = -\log_2 x$, and describe how the graph of $y = \log_{\frac{1}{2}} x$ relates to $y = \log_2 x$.

Solution. Let $y = \log_{\frac{1}{2}} x$. In exponential form:

$$\begin{aligned} \left(\tfrac{1}{2}\right)^y &= x \\ 2^{-y} &= x && \text{since } \tfrac{1}{2} = 2^{-1} \\ -y &= \log_2 x && \text{log form} \\ y &= -\log_2 x \end{aligned}$$

So every $y$-value of $\log_{\frac{1}{2}} x$ is the opposite of the matching $y$-value of $\log_2 x$. The graph of $y = \log_{\frac{1}{2}} x$ is the reflection of $y = \log_2 x$ in the $x$-axis. It’s decreasing, but it still has $x$-intercept $1$ and asymptote $x = 0$.

Check: $\log_{\frac{1}{2}} 8 = -3$ because $\left(\tfrac{1}{2}\right)^{-3} = 8$, and $-\log_2 8 = -3$. ✓

Example 4: Finding the base from a point

The graph of $y = \log_b x$ passes through $(25, 2)$. Find $b$, then find $y$ when $x = \tfrac{1}{5}$.

Solution. Substitute the point and switch to exponential form:

$$2 = \log_b 25 \quad\Rightarrow\quad b^2 = 25 \quad\Rightarrow\quad b = 5$$

(The base must be positive, so $b = -5$ is rejected.) The function is $y = \log_5 x$. When $x = \tfrac{1}{5}$:

$$y = \log_5 \tfrac{1}{5} = -1 \qquad \text{since } 5^{-1} = \tfrac{1}{5}$$

Common mistakes

Giving the logarithmic graph a $y$-intercept of $1$. That’s the exponential graph. The logarithmic graph swaps it: the $x$-intercept is $1$, and there’s no $y$-intercept at all.

Drawing a horizontal asymptote. $y = \log_b x$ has a vertical asymptote, $x = 0$. The graph doesn’t level off; it keeps rising (slowly) without bound, so its range is all real numbers.

Letting the graph cross into $x \le 0$. The domain is $x \gt 0$. Your sketch should never touch or cross the $y$-axis.

Reflecting in the wrong line. The inverse comes from a reflection in $y = x$, not in the $x$-axis or $y$-axis. Swap the coordinates of each point to get it right.

Thinking a bigger base means a steeper graph. For $x \gt 1$, it’s the opposite: $\log_{10} x$ is flatter than $\log_2 x$, because $10$ needs a smaller exponent than $2$ to reach the same number.

Practice

1. (Warm-up) For $y = \log_5 x$, state the domain, range, $x$-intercept, and the equation of the asymptote.

Solution

Domain $\{x \in \mathbb{R} \mid x \gt 0\}$; range $\{y \in \mathbb{R}\}$; $x$-intercept $1$; vertical asymptote $x = 0$.

2. (Warm-up) Write the inverse of each function.

• (a) $y = 7^x$
• (b) $y = \log_9 x$

Solution

(a) $y = \log_7 x$

(b) $y = 9^x$

3. (Warm-up) Find the points on $y = \log_4 x$ for $x = \tfrac{1}{16}$, $\tfrac{1}{4}$, $1$, $4$, $16$, and $64$.

Solution

$\left(\tfrac{1}{16}, -2\right)$, $\left(\tfrac{1}{4}, -1\right)$, $(1, 0)$, $(4, 1)$, $(16, 2)$, $(64, 3)$.

4. (Core) Sketch $y = \log_{\frac{1}{3}} x$ using at least four points. Is it increasing or decreasing? State its domain, range, intercept, and asymptote.

Solution

Swap the points of $y = \left(\tfrac{1}{3}\right)^x$, which are $(-2, 9)$, $(-1, 3)$, $(0, 1)$, $\left(1, \tfrac{1}{3}\right)$, $\left(2, \tfrac{1}{9}\right)$. The points on $y = \log_{\frac{1}{3}} x$ are:

$$(9, -2), \quad (3, -1), \quad (1, 0), \quad \left(\tfrac{1}{3}, 1\right), \quad \left(\tfrac{1}{9}, 2\right)$$

The graph is decreasing: high near the $y$-axis, falling through $(1, 0)$, and continuing slowly downward.

Domain $\{x \in \mathbb{R} \mid x \gt 0\}$; range $\{y \in \mathbb{R}\}$; $x$-intercept $1$; vertical asymptote $x = 0$.

5. (Core) Let $f(x) = \log_2 x$.

• (a) Evaluate $f(32)$ and $f\left(\tfrac{1}{8}\right)$.
• (b) Solve $f(x) = 6$.
• (c) Find $f^{-1}(x)$ and evaluate $f^{-1}(-1)$.

Solution

(a) $f(32) = \log_2 32 = 5$ and $f\left(\tfrac{1}{8}\right) = \log_2 \tfrac{1}{8} = -3$.

(b) $\log_2 x = 6$ means $x = 2^6 = 64$.

(c) $f^{-1}(x) = 2^x$, so $f^{-1}(-1) = 2^{-1} = \tfrac{1}{2}$. Check: $f\left(\tfrac{1}{2}\right) = -1$. ✓

6. (Core) The graph of $y = \log_b x$ passes through $\left(8, \tfrac{3}{2}\right)$. Find $b$, and find $y$ when $x = 2$.

Solution

$b^{\frac{3}{2}} = 8$. Raise both sides to the power $\tfrac{2}{3}$: $b = 8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 4$.

So $y = \log_4 x$, and $\log_4 2 = \tfrac{1}{2}$ because $4^{\frac{1}{2}} = 2$.

7. (Core) Without a calculator, put $\log_2 20$, $\log_3 20$, and $\log_5 20$ in order from largest to smallest. Explain your reasoning.

Solution

$\log_2 20$ is between $4$ and $5$ (since $2^4 = 16$ and $2^5 = 32$). $\log_3 20$ is between $2$ and $3$ (since $9 \lt 20 \lt 27$). $\log_5 20$ is between $1$ and $2$ (since $5 \lt 20 \lt 25$).

So $\log_2 20 \gt \log_3 20 \gt \log_5 20$. A larger base needs a smaller exponent to reach $20$.

8. (Challenge) Explain, using key features of the graphs, why the inverse of $y = 2^x$ is a function but the inverse of $y = x^2$ is not.

Solution

The graph of $y = 2^x$ is always increasing, so every horizontal line crosses it at most once: each output comes from exactly one input. When you reflect it in $y = x$, every vertical line crosses the new graph at most once, so $y = \log_2 x$ is a function.

The graph of $y = x^2$ is a parabola that decreases and then increases. The horizontal line $y = 4$ crosses it twice, at $x = -2$ and $x = 2$. After reflecting, the vertical line $x = 4$ crosses the inverse twice, at $y = -2$ and $y = 2$, so the inverse is not a function.

9. (Challenge) Show that $\log_8 x = \tfrac{1}{3}\log_2 x$ for all $x \gt 0$. What transformation takes the graph of $y = \log_2 x$ to the graph of $y = \log_8 x$?

Solution

Let $y = \log_8 x$. Then:

$$\begin{aligned} 8^y &= x \\ \left(2^3\right)^y &= x \\ 2^{3y} &= x \\ 3y &= \log_2 x && \text{log form} \\ y &= \tfrac{1}{3}\log_2 x \end{aligned}$$

So $y = \log_8 x$ is a vertical compression of $y = \log_2 x$ by a factor of $\tfrac{1}{3}$. Check: $\log_8 64 = 2$ and $\tfrac{1}{3}\log_2 64 = \tfrac{1}{3}(6) = 2$. ✓