Exponential and logarithmic inequalities

An exponential inequality is an exponential equation with an inequality sign in place of the equals sign. Like this one:

$$2^x>8$$

The method starts just like it does for an equation. We write eight as a power of two:

$$2^x>2^3$$

For the equation 2^x = 2³ we would now compare the exponents and get x = 3. Why are we allowed to? The function 2^x is increasing: the bigger the exponent, the bigger the power. So it never takes the same value twice, and two powers of two are equal only when their exponents are equal. The same idea helps with the inequality too, we just also have to decide which way the sign points. Two to the power of something is greater than two cubed exactly when that something is greater than three:

$$x>3$$

The solution is the interval (3, ∞). It checks out: 2⁴ = 16 is more than eight, but 2² = 4 is not. In the picture, it is the part of the x-axis above which the red graph is higher than the green line.

An inequality with a base smaller than one turns out differently, for example ${\left(\frac{1}{2}\right)}^x>8$. The function ${\left(\frac{1}{2}\right)}^x$ is decreasing: the bigger the exponent, the more times we multiply by one half, and the smaller the number we get. In the picture it is blue:

It rises above the line y = 8 only to the left of −3, so the inequality ${\left(\frac{1}{2}\right)}^x>8$ has the solution x < −3. The sign has flipped. The whole trick of exponential inequalities lies in this rule:

• base greater than one: the function is increasing and the inequality sign stays,
• base between zero and one: the function is decreasing and the sign flips.

A base smaller than one

Let’s tackle the inequality

$${\left(\frac{1}{3}\right)}^x\ge 9.$$

We have to write nine as a power of one third. One third to the power of minus one gives three, and to the power of minus two gives nine:

$${\left(\frac{1}{3}\right)}^x\ge {\left(\frac{1}{3}\right)}^{-2}$$

The bases match, but the base $\frac{1}{3}$ is smaller than one. So we compare the exponents and flip the sign:

$$x\le -2$$

The solution is the interval (−∞, −2]. Let’s check with x = −3: ${\left(\frac{1}{3}\right)}^{-3}=3^3=27$, which really is more than nine.

The most common mistake in these questions is: forgetting to flip the sign. Anyone who writes x ≥ −2 would have to accept, say, zero as a solution. But ${\left(\frac{1}{3}\right)}^0=1$, and one is less than nine.

If you don’t like working with fractions in the base, you can rewrite $\frac{1}{3}$ as 3⁻¹. The inequality then reads 3^−x ≥ 3², so −x ≥ 2. This time the base is greater than one and the sign stays, but when we divide by minus one it flips, following the rules for rearranging inequalities. We get x ≤ −2 again. The sign flips either way.

When the bases can’t be made the same

In the inequality 3^x < 5 we can’t write five as a power of three with a whole-number exponent. We deal with it just as we did for equations: we take logarithms of both sides, this time to base three. That logarithm is increasing, so the sign doesn’t change: log₃ 3^x < log₃ 5, so x < log₃ 5. We can happily leave the answer like that. The approximate size of the number is revealed by the inequality 3 < 5 < 9, so log₃ 5 lies between log₃ 3 = 1 and log₃ 9 = 2.

Logarithmic inequalities

Logarithmic inequalities follow the same rule about the base, and there is one more restriction, which we know from logarithmic equations: the argument of a logarithm must be positive. Let’s try it on the inequality

$${\log }_2(x-1)<3.$$

First the restriction: x − 1>0, so x>1. Then we write three as a logarithm to base two, because 2³ = 8:

$${\log }_2(x-1)<{\log }_{2}8$$

The base two is greater than one, the logarithm is increasing and the sign stays:

$$\begin{array}{rcl}x-1& <& 8/+1\\ x& <& 9\end{array}$$

Finally, we combine the result with the restriction. We need x>1 and at the same time x<9, so the solution is the interval (1, 9). To check, we take x = 5: log₂ 4 = 2, which really is less than three.

The picture shows the solution: the red graph is below the green line for every x from one to nine. To the left of one there is no graph at all, because the logarithm does not exist there. That is the restriction x>1.

The mistake here is a different one from the exponential inequalities. Anyone who forgets the restriction will write the answer x<9. But that includes zero as well, and log₂(0 − 1) = log₂(−1) does not exist. With an equation, a forgotten restriction would be exposed by the check, but with an inequality infinitely many numbers come out and we can’t substitute them all. That is why we always write the restriction down right at the start.

The last example brings both traps together:

$${\log }_{\frac{1}{2}}x>2$$

The restriction is x>0. We write two as a logarithm to base one half, because ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$:

$${\log }_{\frac{1}{2}}x>{\log }_{\frac{1}{2}}\frac{1}{4}$$

The base is smaller than one, so the sign flips for a second time:

$$x<\frac{1}{4}$$

With the restriction x>0 we get the solution $(0,\frac{1}{4})$. Let’s check, say, $x=\frac{1}{8}$: one half cubed is one eighth, so ${\log }_{\frac{1}{2}}\frac{1}{8}=3$, and that really is more than two.