# How to Solve Logarithmic & Exponential Inequalities Video

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• 0:00 Background Example
• 0:34 Exponential Inequalities
• 3:16 Logarithmic Inequalities
• 8:23 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we will explore logarithmic and exponential inequalities while showing how they relate to value calculations. We will show examples of several different types of these inequalities.

## Background Example

After years of saving, Samantha is finally buying a new car. However, her main concern is her car will lose too much value in the long run. With an annual depreciation of 16.8%, Samantha is wondering how many years does the car have before the car's value dips below half of the retail price? This is an exponential inequality problem. After exploring exponential and logarithmic inequalities, this lesson responds to Samantha's concerns.

## Exponential Inequalities

Let's first lay out some concrete definitions that distinguish exponential and logarithmic inequalities. Simply put, exponential inequalities have exponents while logarithmic inequalities have logarithms. Both use one of four types of inequalities. In words: greater than, greater than or equal to, less than, and less than or equal to.

Before showing a solution method, let's solve a similar math problem by starting with the answer. Sound interesting? What is 25? Answer: 2(2)(2)(2)(2) = 32. That's the answer, but what's the question?

The question is: find x so 2x > 32.

Since 25 is 32, any x greater than 5 will work.

Now for a solution method:

Step 1: Replace the inequality with an equal sign.

From 2x > 32 write

Step 2: With exponents, use logarithms.

Take the log:

Step 3: Solve.

From the property, log ax = x log a, bring x out in front:

Divide by log 2:

Step 4: Evaluate.

Using a calculator, log 32 = 1.505 and log 2 = 0.301.Thus, x = 1.505 / 0.301 = 5.

Step 5: Determine the domain.

What values of x give 2x > 32? Try a number greater than 5, like x = 6. Then, 26 = 64 is greater than 32. So, numbers larger than 5 work. Just to be sure, how about a number less than 5, like x = 4? Well, 24 = 16 is not greater than 32.

Thus, x > 5 satisfies the inequality 2x > 32.

Step 6: (an optional step) Plot.

On a number line show the solution values for x:

A circle shows the missing value at x = 5 and the values for x extend to positive infinity. The value 5 is not included because the original inequality was strictly 'greater than' and not 'greater than or equal to.'

## Logarithmic Inequalities

Logarithms and exponentials are inverse operations. In other words, one operation undoes the other. For example, 102 = 100 and log 100 = 2. Also, 10log x = x. Note: log is base 10.

Find the x values for log x < 2.

Step 1: Replace the inequality with an equal sign.

From log x < 2, write

Step 2: With a logarithm, raise to the power of the base.

Raise to the power of 10:

Step 3: Solve.

Using the inverse property, alogx = x:

Step 4: Evaluate.

Step 5: Determine the domain.

x = 101, log 101 = 2.004 is greater than 2 and not in the domain. x = 99, log 99 = 1.995 is less than 2 and in the domain. So, our conclusion should be x < 100. Unlike exponentials, there is a special consideration with logarithms: log x is valid for x greater than zero. Thus, x < 100 and x > 0. Combining these inequalities, we get: 0 < x < 100.

Step 6: Plot.

Circles show the missing values at x = 0 and x = 100. The 0 is not included because log(0) is not defined and the 100 isn't included because the original inequality problem stated less than' and not 'less than or equal to.'

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