Graphs of logarithms and log inequalities

Now that log­a­rithms are defined, we can treat $y = \log_a x$ as a func­tion and draw its graph. The graph looks very dif­fer­ent for a base big­ger than $1$ and a base between $0$ and $1$, and that dif­fer­ence is the whole secret of log­a­rith­mic inequal­i­ties: whether $\log_a x$ is pos­i­tive or neg­a­tive depends on where both $a$ and $x$ sit com­pared with $1$.

The graph of $y = \log_2 x$#

Make a table using the laws of log­a­rithms.

• $x = 1$: $y = \log_2 1 = 0$, giv­ing $(1, 0)$.
• $x = 2$: $y = \log_2 2 = 1$, giv­ing $(2, 1)$.
• $x = 4$: $y = \log_2 4 = \log_2 2^2 = 2\log_2 2 = 2(1) = 2$, giv­ing $(4, 2)$.
• $\displaystyle x = \frac{1}{2}$: $y = \log_2 2^{-1} = (-1)\log_2 2 = -1$, giv­ing $\displaystyle \left(\frac{1}{2}, -1\right)$.
• $\displaystyle x = \frac{1}{4}$: $y = -2$, giv­ing $\displaystyle \left(\frac{1}{4}, -2\right)$.

Plot the points and join them with a smooth curve. The curve rises from left to right: it is an increas­ing curve. It crosses the $x$-axis at $(1, 0)$. As $x$ gets close to $0$ the curve drops steeply down­wards, hug­ging the $y$-axis but never touch­ing it, because $\log_2 0$ does not exist. To the right it keeps ris­ing, but more and more slowly.

The graph of $y = \log_{1/2} x$#

• $x = 1$: $y = \log_{1/2} 1 = 0$, giv­ing $(1, 0)$.
• $x = 2$: $\displaystyle y = \log_{2^{-1}} 2^1 = \frac{1}{-1}\log_2 2 = -1$, giv­ing $(2, -1)$.
• $x = 4$: $y = -2$, giv­ing $(4, -2)$.
• $\displaystyle x = \frac{1}{2}$: $\displaystyle y = \log_{1/2} \frac{1}{2} = 1$, giv­ing $\displaystyle \left(\frac{1}{2}, 1\right)$.
• $\displaystyle x = \frac{1}{4}$: $y = 2$, giv­ing $\displaystyle \left(\frac{1}{4}, 2\right)$.

This time the curve falls from left to right: it is a decreas­ing curve. It is high up near the $y$-axis, crosses the $x$-axis at $(1, 0)$, and goes below the axis for $x > 1$. It is the mir­ror image of $y = \log_2 x$ in the $x$-axis.

The two shapes in gen­eral#

• If $a > 1$, the graph of $y = \log_a x$ is increas­ing, passes through $(1, 0)$, and drops steeply near the $y$-axis. It is below the $x$-axis for $0 < x < 1$ and above it for $x > 1$.
• If $0 < a < 1$, the graph of $y = \log_a x$ is decreas­ing, passes through $(1, 0)$, and rises steeply near the $y$-axis. It is above the $x$-axis for $0 < x < 1$ and below it for $x > 1$.

In both cases the domain is $(0, \infty)$ and the range is $\mathbb{R}$, and every graph passes through $(1, 0)$, because $\log_a 1 = 0$ for every allowed base.

Signs of a log­a­rithm#

Read­ing the two graphs gives four rules.

1. $\log_a x > 0 \iff$ either $a > 1$ and $x > 1$, or $0 < a < 1$ and $0 < x < 1$.
2. $\log_a x \geq 0 \iff$ either $a > 1$ and $x \geq 1$, or $0 < a < 1$ and $0 < x \leq 1$.
3. $\log_a x < 0 \iff$ either $a > 1$ and $0 < x < 1$, or $0 < a < 1$ and $x > 1$.
4. $\log_a x \leq 0 \iff$ either $a > 1$ and $0 < x \leq 1$, or $0 < a < 1$ and $x \geq 1$.

An easy way to hold this in your head: the log­a­rithm is pos­i­tive when the base and the num­ber are on the same side of $1$, and neg­a­tive when they are on oppo­site sides.

Exam­ple 1#

State whether each log­a­rithm is pos­i­tive or neg­a­tive.

Log­a­rith­mic inequal­i­ties#

To solve an inequal­ity such as $\log_a (\text{expression}) > 0$, use the sign rules to turn it into an ordi­nary inequal­ity. Remem­ber also that the expres­sion inside the log­a­rithm must be pos­i­tive.

Exam­ple 2#

If $\log_{10} (x - 2) > 0$, find the val­ues of $x$.

The base $10$ is greater than $1$, so the log­a­rithm is pos­i­tive exactly when the num­ber is greater than $1$:

$$\begin{aligned} x - 2 &> 1 \\ x &> 3 \end{aligned}$$

So $x \in (3, \infty)$. The con­di­tion $x - 2 > 0$ is auto­mat­i­cally met, since $x > 3$.

Exam­ple 3#

Solve $\log_{1/3} (x - 1) > 0$.

The base $\displaystyle \frac{1}{3}$ is less than $1$, so the log­a­rithm is pos­i­tive exactly when the num­ber is between $0$ and $1$:

$$0 < x - 1 < 1 \implies 1 < x < 2$$

So $x \in (1, 2)$.

Exam­ple 4#

Solve $\log_2 (2x - 3) \leq 0$.

The base is greater than $1$, so we need the num­ber to be pos­i­tive and at most $1$:

$$\displaystyle 0 < 2x - 3 \leq 1 \implies 3 < 2x \leq 4 \implies \frac{3}{2} < x \leq 2$$

So $\displaystyle x \in \left(\frac{3}{2}, 2\right]$. The left end is open because $\log_2 0$ is not defined.

Prac­tice#

1. Is $\log_5 7$ pos­i­tive or neg­a­tive?
2. Is $\displaystyle \log_5 \frac{1}{7}$ pos­i­tive or neg­a­tive?
3. Is $\log_{0.2} 0.7$ pos­i­tive or neg­a­tive?
4. Is $\log_{0.2} 7$ pos­i­tive or neg­a­tive?
5. What is the value of $\log_3 1$?
6. Find $y = \log_3 x$ for $\displaystyle x = \frac{1}{9}, \frac{1}{3}, 1, 3, 9$, and say whether the graph rises or falls.
7. Solve $\log_2 (x + 1) > 0$.
8. Solve $\log_{1/2} (x - 3) > 0$.
9. Solve $\log_3 (2x - 1) \leq 0$.
10. Solve $\log_{0.5} (x + 2) \leq 0$.
11. With­out a cal­cu­la­tor, say which is greater: $\log_2 5$ or $\displaystyle \log_3 \frac{1}{2}$.

Answers#