Lesson Objectives

- Demonstrate an understanding of how to solve exponential equations
- Demonstrate an understanding of how to solve logarithmic equations
- Learn how to solve exponential and logarithmic inequalities

## How to Solve Exponential and Logarithmic Inequalities

In this lesson, we will learn how to solve exponential and logarithmic inequalities. Throughout our course, we have encountered various types of inequalities and explored different strategies to solve them. Here, we will provide a general method that works for all cases. Note, the method given here is usually not the fastest but is easy to follow and removes some complexity from other methods.

5) Test a point in each interval. We can use the inequality from 2.2: $$2^{7x + 12}- 2^6 > 0$$ Pick anything less than -6/7 and plug in for x. We just want to see if the left side is greater than 0. Then pick anything greater than -6/7 and plug in for x. Again, we just want to see if the left side is greater than 0.

6) Our solution for x will be anything that is greater than -6/7. $$x > -\frac{6}{7}$$ Interval Notation: $$\left(-\frac{6}{7}, \infty\right)$$ Graphing the Interval:

Example #2: Solve each inequality. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Let's begin with our given procedure.

1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract (1/2)

5) Test a point in each interval. We can use the inequality from 2.2: $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ Pick anything less than 3 and plug in for x. We just want to see if the left side is greater than 0. Then pick anything greater than 3 and plug in for x. Again, we just want to see if the left side is greater than 0.

6) Our solution for x will be anything that is less than 3. $$x < 3$$ Interval Notation: $$\left(-\infty, 3\right)$$ Graphing the Interval: Using the faster approach we must modify our formula, otherwise, we will get the wrong answer. $$0 < a < 1 \hspace{.1em}\text{and}\hspace{.1em}x > y$$ then: $$a^x < a^y$$ Notice the flipping of the sign here. If you don't do this, you will get the wrong answer which is why the general strategy may be preferred. There is just less to remember and it always works.

The reasoning behind flipping the sign is very straightforward. Let's just think about something like 1/2. $$\left(\frac{1}{2}\right)^2=\frac{1}{4}$$ $$\left(\frac{1}{2}\right)^4=\frac{1}{16}$$ $$\left(\frac{1}{2}\right)^6=\frac{1}{64}$$ As the exponent is increasing, the result is decreasing. So if x > y, meaning the exponent is larger and we have some base that is between 0 and 1 (not including either), then a larger exponent results in a smaller number and so the sign must be flipped. Let's rework the problem using this strategy. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Simplify and write each side with a common base. $$\left(\frac{1}{2}\right)^{3x + 3}> \left(\frac{1}{2}\right)^{12}$$ You can't just set 3x + 3 > 12 and solve. You must flip the direction of the inequality symbol. $$3x + 3 > 12{\color{red}✗}\text{wrong!}$$ $$3x + 3 < 12{\color{green}✓}\text{correct!}$$ Let's solve the inequality: $$3x + 3 < 12$$ Subtract 3 from each side: $$3x < 9$$ Divide both sides by 3: $$x < 3$$ Now, we see the problems that can arise when using the shorter method. From here on out, we will just stick with the general strategy given.

5) Test a point in each interval. We can use the inequality from 2.2: $$7^{4x + 2}- 14 ≤ 0$$ We can just use a simple approximation of -0.16. We just need to check values on each side, so we can pick something for x such as -1, which is to the left of -0.16, and 0, which is to the right of -0.16. Again, we just want to see if the left side is less than 0.

6) Our solution for x will be anything that is less than or equal to about -0.16. $$x ≤ \frac{\log_7(14) - 2}{4}$$ Interval Notation: $$\left(-\infty, \frac{\log_7(14) - 2}{4}\right]$$ Graphing the Interval:

Example #4: Solve each inequality. $$\log_6(x + 1) ≥ \log_6(4 - 2x)$$ 1) What is the domain? Here, the process is more involved. Remember that the argument of a logarithm is a positive real number. $$x + 1 > 0$$ $$\text{and}$$ $$4 - 2x > 0$$ Both arguments must be positive, so we want the intersection of the two inequalities.

First inequality: $$x + 1 > 0$$ Subtract 1 away from each side: $$x > -1$$ Second inequality: $$4 - 2x > 0$$ Subtract 4 away from each side: $$-2x > -4$$ Divide both sides by -2, don't forget to flip the direction of the inequality symbol: $$x < 2$$ Now we have the following: $$x > -1$$ $$\text{and}$$ $$x < 2$$ We can use a three-part inequality to write the domain: $$\text{domain}: \{x |{-}1 < x < 2\}$$ 2.1) Change the right side into 0. Subtract log

5) Test a point in each interval. We can use the inequality from 2.2: $$\log_6\left(\frac{x + 1}{-2x + 4}\right) ≥ 0$$

6) Our solution for x will be anything that is between 1 and 2. We also need to include 1 here because we have a non-strict inequality. $$1 ≤ x < 2$$ Interval Notation: $$[1, 2)$$ Graphing the Interval: For this type of problem, it can be helpful to look at the graph of the original inequality. $$\log_6(x + 1) ≥ \log_6(4 - 2x)$$ $$\log_6(x + 1) - \log_6(4 - 2x)≥ 0$$ Let's graph this as: $$y=\log_6(x + 1) - \log_6(4 - 2x)$$ Here we are looking for the x-values for which the y-values are greater than or equal to 0. We can see from the graph below, this occurs when x is between 1 and 2, where 1 is included and 2 is excluded. Example #5: Solve each inequality. $$\log_4(x^2 + 8) - \log_4(6) < \log_4(2)$$ 1) What is the domain? Here, the process is more involved. Remember that the argument of a logarithm is a positive real number. $$x^2 + 8 > 0$$ The solution to this inequality is all real numbers, therefore, our domain will include all real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract log

5) Test a point in each interval. We can use the inequality from 2.2: $$\log_4\left(\frac{x^2 + 8}{12}\right) < 0$$

6) Our solution for x will be anything that is between -2 and 2 with both being excluded. $$-2 < x < 2$$ Interval Notation: $$(-2, 2)$$ Graphing the Interval: Again, just like in the last example, a graph can be helpful. $$\log_4(x^2 + 8) - \log_4(6) < \log_4(2)$$ $$\log_4(x^2 + 8) - \log_4(6) - \log_4(2) < 0$$ Let's graph this as: $$y=\log_4(x^2 + 8) - \log_4(6) - \log_4(2)$$ Here we are looking for the x-values for which the y-values are less than 0. We can see from the graph below, this occurs when x is between -2 and 2, where both are excluded.

### General Method for Solving an Exponential or Logarithmic Inequality

- Identify the domain
- Remember that the argument for a logarithm is a positive real number

- Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible
- Find the critical values
- Replace the inequality symbol with an equality symbol and solve the resulting equation
- These critical values are where the function changes sign
- Critical values also occur where the function is undefined, found in step 1

- Split the number line up into intervals based on the critical values
- Solution(s) to our related equation
- Any undefined points

- Test a point in each interval
- Use a sign chart to keep track of the sign of the function
- Consider the critical values separately

- The solution is the union of all intervals where the function is positive or negative, depending on the inequality sign given in the problem

$$\left(-\infty, -\frac{6}{7}\right)$$ | $$\left(-\frac{6}{7}, \infty\right)$$ |
---|

$$\left(-\infty, -\frac{6}{7}\right)$$ | $$\left(-\frac{6}{7}, \infty\right)$$ |
---|---|

(-) | (+) |

### A faster approach for exponential inequalities with like bases

Seems like quite a bit of work for such a simple problem. You may be asking, is there a faster method? Yes, but it involves a bit of a rule that must be memorized. $$a > 1 \hspace{.1em}\text{and}\hspace{.1em}x > y$$ then: $$a^x > a^y$$ Notice that this is true for a is greater than 1. We will see in a moment why this may cause confusion and lead to a wrong answer when the base is between 0 and 1 (not including either). Let's rework the problem using this strategy. $$32^{3x}\cdot \left(\frac{1}{16}\right)^{2x - 3}> 64$$ Simplify and write each side with a common base. $$2^{7x + 12}> 2^{6}$$ You can just set 7x + 12 > 6 and solve the simple linear inequality. $$7x + 12 > 6$$ Subtract 12 away from each side: $$7x > -6$$ Divide both sides by 7: $$x > -\frac{6}{7}$$ Of course this is much quicker but let's see the issue that will come up on the next problem.Example #2: Solve each inequality. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Let's begin with our given procedure.

1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract (1/2)

^{12}away from each side: $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ 2.2) Simplify the left side. $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}> 0$$ 3) Find the critical values. Since the domain is the set of real numbers, we just need to solve the resulting equation (replace the inequality symbol with an equality symbol). $$\left(\frac{1}{2}\right)^{3x + 3}- \left(\frac{1}{2}\right)^{12}=0$$ $$\left(\frac{1}{2}\right)^{3x + 3}=\left(\frac{1}{2}\right)^{12}$$ Since the base is 1/2 on each side, we can just set the exponents equal and solve. $$3x + 3=12$$ Subtract 3 away from each side: $$3x=9$$ Divide both sides by 3: $$x=3$$ 4) Split the number line up into intervals based on the critical values.$$\left(-\infty, 3\right)$$ | $$\left(3, \infty\right)$$ |
---|

$$\left(-\infty, 3\right)$$ | $$\left(3, \infty\right)$$ |
---|---|

(+) | (-) |

The reasoning behind flipping the sign is very straightforward. Let's just think about something like 1/2. $$\left(\frac{1}{2}\right)^2=\frac{1}{4}$$ $$\left(\frac{1}{2}\right)^4=\frac{1}{16}$$ $$\left(\frac{1}{2}\right)^6=\frac{1}{64}$$ As the exponent is increasing, the result is decreasing. So if x > y, meaning the exponent is larger and we have some base that is between 0 and 1 (not including either), then a larger exponent results in a smaller number and so the sign must be flipped. Let's rework the problem using this strategy. $$\left(\frac{1}{2}\right)^x \cdot \left(\frac{1}{2}\right)^{2x + 3}> \left(\frac{1}{2}\right)^{12}$$ Simplify and write each side with a common base. $$\left(\frac{1}{2}\right)^{3x + 3}> \left(\frac{1}{2}\right)^{12}$$ You can't just set 3x + 3 > 12 and solve. You must flip the direction of the inequality symbol. $$3x + 3 > 12{\color{red}✗}\text{wrong!}$$ $$3x + 3 < 12{\color{green}✓}\text{correct!}$$ Let's solve the inequality: $$3x + 3 < 12$$ Subtract 3 from each side: $$3x < 9$$ Divide both sides by 3: $$x < 3$$ Now, we see the problems that can arise when using the shorter method. From here on out, we will just stick with the general strategy given.

### Solving Exponential Inequalities with Unlike Bases

Example #3: Solve each inequality. $$3 \cdot 7^{4x + 2}+ 9 ≤ 51$$ 1) What is the domain? The set of real numbers. $$\text{domain}: \left\{x | x ∈ \mathbb{R}\right\}$$ 2.1) Change the right side into 0. Subtract 51 away from each side: $$3 \cdot 7^{4x + 2}- 42 ≤ 0$$ 2.2) Simplify the left side. To do this, let's divide each side by 3: $$7^{4x + 2}- 14 ≤ 0$$ 3) Find the critical values. Since the domain is the set of real numbers, we just need to solve the resulting equation (replace the inequality symbol with an equality symbol). $$7^{4x + 2}- 14=0$$ Add 14 to each side: $$7^{4x + 2}=14$$ Take log base 7 of each side: $$\log_7(7^{4x + 2})=\log_7(14)$$ $$4x + 2=\log_7(14)$$ Subtract 2 away from each side: $$4x=\log_7(14) - 2$$ Divide both sides by 4: $$x=\frac{\log_7(14) - 2}{4}$$ 4) Split the number line up into intervals based on the critical values.$$\left(-\infty, \frac{\log_7(14) - 2}{4}\right)$$ | $$\left(\frac{\log_7(14) - 2}{4}, \infty\right)$$ |
---|

$$\left(-\infty, \frac{\log_7(14) - 2}{4}\right)$$ | $$\left(\frac{\log_7(14) - 2}{4}, \infty\right)$$ |
---|---|

(-) | (+) |

### Solving Logarithmic Inequalities

Let's look at a few problems with logarithms.Example #4: Solve each inequality. $$\log_6(x + 1) ≥ \log_6(4 - 2x)$$ 1) What is the domain? Here, the process is more involved. Remember that the argument of a logarithm is a positive real number. $$x + 1 > 0$$ $$\text{and}$$ $$4 - 2x > 0$$ Both arguments must be positive, so we want the intersection of the two inequalities.

First inequality: $$x + 1 > 0$$ Subtract 1 away from each side: $$x > -1$$ Second inequality: $$4 - 2x > 0$$ Subtract 4 away from each side: $$-2x > -4$$ Divide both sides by -2, don't forget to flip the direction of the inequality symbol: $$x < 2$$ Now we have the following: $$x > -1$$ $$\text{and}$$ $$x < 2$$ We can use a three-part inequality to write the domain: $$\text{domain}: \{x |{-}1 < x < 2\}$$ 2.1) Change the right side into 0. Subtract log

_{6}(4 - 2x) away from each side: $$\log_6(x + 1) - \log_6(4 - 2x) ≥ 0$$ 2.2) Simplify the left side. To do this, let's use our quotient rule for logarithms. $$\log_6\left(\frac{x + 1}{-2x + 4}\right) ≥ 0$$ 3) Find the critical values. Let's solve the following equation. $$\log_6(x + 1)=\log_6(4 - 2x)$$ $$x + 1=4 - 2x$$ Add 2x to each side: $$3x + 1=4$$ Subtract 1 away from each side: $$3x=3$$ Divide both sides by 3: $$x=1$$ 4) Split the number line up into intervals based on the critical values. We will use our domain from part 1 along with our solution from the equation in part 3.$$\left(-\infty, -1\right)$$ | $$\left(-1, 1\right)$$ | $$\left(1, 2\right)$$ | $$\left(2, \infty\right)$$ |
---|

$$\left(-\infty, -1\right)$$ | $$\left(-1, 1\right)$$ | $$\left(1, 2\right)$$ | $$\left(2, \infty\right)$$ |
---|---|---|---|

undefined | (-) | (+) | undefined |

_{4}(2) away from each side: $$\log_4(x^2 + 8) - \log_4(6) - \log_4(2) < 0$$ 2.2) Simplify the left side. To do this, let's use our product and quotient rules for logarithms. $$\log_4(x^2 + 8) - (\log_4(6) + \log_4(2)) < 0$$ $$\log_4(x^2 + 8) - \log_4(12) < 0$$ $$\log_4\left(\frac{x^2 + 8}{12}\right) < 0$$ 3) Find the critical values. Let's solve the following equation. $$\log_4\left(\frac{x^2 + 8}{12}\right)=0$$ $$4^{0}=\frac{x^2 + 8}{12}$$ Switch sides, write 4^{0}as 1: $$\frac{x^2 + 8}{12}=1$$ Multiply both sides by 12: $$x^2 + 8=12$$ Subtract 8 away from each side: $$x^2=4$$ Square root property: $$x=\pm 2$$ 4) Split the number line up into intervals based on the critical values. We will use our domain from part 1 along with our solution from the equation in part 3.$$\left(-\infty, -2\right)$$ | $$\left(-2, 2\right)$$ | $$\left(2, \infty\right)$$ |
---|

$$\left(-\infty, -2\right)$$ | $$\left(-2, 2\right)$$ | $$\left(2, \infty\right)$$ |
---|---|---|

(+) | (-) | (+) |

#### Skills Check:

Example #1

Solve each inequality.

$$\frac{27^{x - 1}}{243^{3 - 3x}}> 27^x$$

Please choose the best answer.

A

$$x > \frac{6}{5}$$

B

$$x < \frac{6}{5}$$

C

$$\text{No Solution}$$

D

$$-\frac{2}{3}< x < \frac{2}{3}$$

E

$$x > -1$$

Example #2

Solve each inequality.

$$\log_2(x^2 - 8x) < \log_2(3x - 18)$$

A

$$x < 9$$

B

$$x > 9$$

C

$$8 < x < 9$$

D

$$-\frac{1}{3}< x < \frac{1}{3}$$

E

$$\text{No Solution}$$

Example #3

Solve each inequality.

$$\log_3(x - 4) - \log_3(x + 3) > 2$$

Please choose the best answer.

A

$$x > -\frac{2}{5}$$

B

$$\text{No Solution}$$

C

$$x > \frac{1}{27}$$

D

$$-3 < x < 3$$

E

$$x > -2$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test