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Exponential and Logarithmic Equations and Inequalities

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, this topic is about using algebra "superpowers"—exponents and logarithms—to solve for unknown variables and find inverse functions.

Why this matters

Imagine you’re saving up for a down payment on your first car. You have $4,000 in a savings account that earns 5% interest, compounded annually. You need to reach $5,000. The question is, how long will it take?

You can set up the compound interest formula: 5000 = 4000(1.05)^t. The variable you need to find, t for time, is stuck up in the exponent. How do you get it down? You can’t just divide by 1.05. This isn't a simple algebra problem from last year.

This is exactly the kind of question we'll tackle today. We're learning the techniques to solve for variables when they're trapped in exponents or inside logarithms. These tools are essential not just for finance, but for science, computer science, and engineering. Let's get started.

Growth of $4000 at 5% annual interest, showing time to reach $5000.

Concept overview

flowchart TD
    A[Start: Solve Equation] --> B{Type?};
    B --> C[Exponential: b^x = c];
    B --> D[Logarithmic: log(x) = c];
    C --> E{Can you get a common base?};
    E --> F[Yes: Set exponents equal and solve];
    E --> G[No: Take log of both sides and solve];
    D --> H{Can you condense to log(A) = log(C)?};
    H --> I[Yes: Set A = C and solve];
    H --> J[No: Convert to exponential form and solve];
    I --> K((Must check for extraneous solutions!));
    J --> K;
    F --> L[Final Answer];
    G --> L;
    K --> L;

Core explanation

Alright, let's dive in. Think of exponential and logarithmic functions as two sides of the same coin. They are inverses of each other, which means one undoes the other. This relationship is the key to everything we're doing today.

Solving Exponential Equations

When you have an equation with the variable in the exponent, your goal is to get it out.

Strategy 1: Find a Common Base

If you can rewrite both sides of the equation to have the same base, you're in great shape.

For example, solve 2^(x+1) = 8. You know that 8 is 2^3. So you can rewrite the equation: 2^(x+1) = 2^3

Since the bases are the same (and they're not 1, 0, or -1), the exponents must be equal. x + 1 = 3 x = 2

This is the most direct method, so always look for it first.

Strategy 2: Take the Logarithm of Both Sides

What if you can't find a common base, like in our car savings problem 5000 = 4000(1.05)^t? First, let's isolate the exponential part: 1.25 = (1.05)^t

Now, we can't easily write 1.25 as a power of 1.05. So, we bring in the superpower: logarithms. You can take the log of both sides of any equation (as long as both sides are positive). You can use any base log, but your calculator has buttons for log (base 10) and ln (base e). Let's use ln.

ln(1.25) = ln(1.05^t)

Now, use the power property of logarithms, which lets you bring the exponent down in front: ln(1.25) = t * ln(1.05)

The variable t is finally free! To solve for it, just divide. t = ln(1.25) / ln(1.05)

Using a calculator, t ≈ 4.57. So, it will take about 4.6 years to save for the down payment.

Solving Logarithmic Equations

Solving log equations often involves using log properties to simplify the equation first.

Strategy 1: Use the Definition of a Logarithm

If you have a single logarithm equal to a number, like log_3(x - 5) = 2, you can "exponentiate" both sides. That's a fancy way of saying, convert it from log form to exponential form.

Remember: log_b(A) = C is the same as b^C = A. Applying this here: 3^2 = x - 5 9 = x - 5 x = 14

Strategy 2: Use the One-to-One Property

If you can get your equation into the form log_b(A) = log_b(C), then you can conclude that A = C.

For example: ln(x) + ln(x - 4) = ln(5) First, use the product property to condense the left side: ln(x * (x - 4)) = ln(5)

Now you have the ln(A) = ln(C) form. So: x(x - 4) = 5 x^2 - 4x = 5 x^2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 This gives two potential solutions: x = 5 and x = -1.

  • Check x = 5: ln(5) is fine. ln(5 - 4) = ln(1) is fine. So, x = 5 is a valid solution.
  • Check x = -1: ln(-1) is undefined! You can't take the log of a negative number. So, x = -1 is an extraneous solution. We throw it out.

The only solution is x = 5. Always, always check your answers for log equations.

Finding Inverse Functions

Finding an inverse is like writing down the "undo" instructions for a function. The process is always the same:

  1. Replace f(x) with y.
  2. Swap x and y everywhere in the equation.
  3. Solve the new equation for y.
  4. Replace y with f⁻¹(x).

Inverse of an Exponential Function

Let's find the inverse of f(x) = 2 * 5^(x-1) + 3.

  1. y = 2 * 5^(x-1) + 3
  2. Swap x and y: x = 2 * 5^(y-1) + 3
  3. Solve for y. This is just like the equation solving we did above! Isolate the exponential term first. x - 3 = 2 * 5^(y-1) (x - 3) / 2 = 5^(y-1) Now, take log_5 of both sides to free the exponent. log_5((x - 3) / 2) = y - 1 Finally, add 1. y = log_5((x - 3) / 2) + 1
  4. f⁻¹(x) = log_5((x - 3) / 2) + 1

Inverse of a Logarithmic Function

Let's find the inverse of g(x) = 4 * ln(x + 2) - 7.

  1. y = 4 * ln(x + 2) - 7
  2. Swap x and y: x = 4 * ln(y + 2) - 7
  3. Solve for y. Isolate the logarithm. x + 7 = 4 * ln(y + 2) (x + 7) / 4 = ln(y + 2) Now, exponentiate both sides using base e to undo the natural log. e^((x + 7) / 4) = y + 2 Finally, subtract 2. y = e^((x + 7) / 4) - 2
  4. g⁻¹(x) = e^((x + 7) / 4) - 2

And just as we expected, the inverse of a logarithmic function is an exponential function.

Visualizing the inverse relationship between exponential and logarithmic functions.

Worked examples

Let's walk through a few problems together. The key is to identify the type of problem first, then apply the right strategy.

Example 1

Solving a Logarithmic Equation with an Extraneous Solution

Problem: Solve the equation log_2(x) + log_2(x - 2) = 3.

Solution Walkthrough:

  1. 1
    Identify the Goal
    We need to solve for x. The x is inside logarithms. Our first step should be to combine these logs into a single logarithm.
  2. 2
    Apply Log Properties
    The sum of two logs with the same base is the log of their product. log_2(x * (x - 2)) = 3 log_2(x^2 - 2x) = 3
  3. 3
    Convert to Exponential Form
    Now we have a single log equal to a number. This is the perfect time to use the definition of a logarithm (log_b(A) = C means b^C = A). 2^3 = x^2 - 2x 8 = x^2 - 2x
  4. 4
    Solve the Resulting Equation
    This is a quadratic equation. Let's get it equal to zero. 0 = x^2 - 2x - 8 Factor the quadratic. We need two numbers that multiply to -8 and add to -2. That's -4 and +2. 0 = (x - 4)(x + 2) This gives us two potential solutions: x = 4 and x = -2.
  5. 5
    Check for Extraneous Solutions
    This is the most critical step for log equations. We must plug our answers back into the original equation: log_2(x) + log_2(x - 2) = 3.
    • Check x = 4: log_2(4) is valid. log_2(4 - 2) = log_2(2) is also valid. So, x = 4 is a real solution.
    • Check x = -2: log_2(-2)... stop right there. You cannot take the logarithm of a negative number. The domain of log_2(x) is x > 0. Therefore, x = -2 is an extraneous solution.

Final Answer: The only solution is x = 4.

Example 2

Finding the Inverse of a Transformed Exponential Function

Problem: Given f(x) = 3 * (1/2)^(x+4) - 5, find the inverse function, f⁻¹(x).

Solution Walkthrough:

  1. Rewrite with y: Start by replacing f(x) with y. y = 3 * (1/2)^(x+4) - 5
  2. Swap x and y: This is the key step to finding an inverse. Every x becomes a y and every y becomes an x. x = 3 * (1/2)^(y+4) - 5
  3. Solve for y: Now, we treat this as an algebra puzzle. Our goal is to isolate y. We undo the operations in reverse order (reverse PEMDAS).
    • Add 5 to both sides: x + 5 = 3 * (1/2)^(y+4)
    • Divide by 3: (x + 5) / 3 = (1/2)^(y+4)
    • Now, to get y out of the exponent, we take the logarithm of both sides. Since the base is 1/2, it's most natural to use log_1/2. log_1/2((x + 5) / 3) = y + 4
    • Finally, subtract 4 from both sides: y = log_1/2((x + 5) / 3) - 4
  4. State the Inverse Function: Replace y with the proper notation f⁻¹(x).

Final Answer: f⁻¹(x) = log_1/2((x + 5) / 3) - 4.

Try it yourself

Ready to try a couple on your own? Don't worry about getting the perfect answer right away. Focus on setting up the problem correctly and identifying the right strategy.

1. Solve the Inequality

Solve 2 * ln(x - 3) > 8.

Hint 1: Start by isolating the logarithm, just like you would in an equation. Hint 2: After you exponentiate, remember that logarithms have a domain. What's the simplest inequality you can write just by looking at ln(x-3)? Your final answer must satisfy both conditions.

2. Find the Inverse

Find the inverse of the function g(x) = e^(2x) - 5.

Hint: Swap x and y. Your goal is to solve for the new y. You'll need to use a logarithm to get y out of the exponent. Which logarithm is the inverse of base e?

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