1. Remember that exponential functions and logarithmic functions are inverses of each other. We will use this property to solve problems.

2. S OLVING E XPONENTIAL E QUATIONS x = 5

3. Solving by Equating Exponents Solve 4 3x = 8 x + 1. ( 2 2 ) 3x = ( 2 3 ) x + 1 6x = 3x + 3 2 6x = 2 3x + 3 2 2 (3x) = 2 3(x + 1) x = 1 The solution is 1. S OLUTION Rewrite each power with base 2. Power of a power property Equate exponents. Solve for x. 4 3 x = 8 x + 1 Write original equation. C HECK Check the solution by substituting it into the original equation. 4 3 1 = 8 1 + 1 64 = 64 Solution checks. Solve for x.

4. Solving by Equating Exponents When it is not convenient to write each side of an exponential equation using the same base, you can solve the equation by taking a logarithm of each side.

5. Taking a Logarithm of Each Side Solve 10 2 x – 3 + 4 = 21. 10 2 x – 3 = 17 log 10 2 x – 3 = log 17 2 x = 3 + 1.23 x = (3 + 1.23 ) 1 2 x  2.115 Use a calculator. 10 2 x – 3 + 4 = 21 S OLUTION Write original equation. Subtract 4 from each side. Add 3 to each side. Multiply each side by. 1 2 Take log base 10 of each side. 2 x – 3 = log 17 log 10 x = x

6. C HECK Taking a Logarithm of Each Side Check the solution algebraically by substituting into the original equation. Solve 10 2 x – 3 + 4 = 21.

7. S OLVING L OGARITHMIC E QUATIONS To solve a logarithmic equation, use this property for logarithms with the same base: For positive numbers b, x, and y where b  1, log b x = log b y if and only if x = y.

8. Use property for logarithms with the same base. 5 x = x + 8 Solving a Logarithmic Equation Solve log 3 (5 x – 1) = log 3 (x + 7). 5 x – 1 = x + 7 x = 2 The solution is 2. S OLUTION Use inverse property with base 3. Add 1 to each side. Solve for x.

9. Use property for logarithms with the same base. 5 x = x + 8 Solving a Logarithmic Equation 5 x – 1 = x + 7 x = 2 S OLUTION log 3 (5 x – 1) = log 3 (x + 7) Write original equation. Add 1 to each side. Solve for x. C HECK Check the solution by substituting it into the original equation. log 3 (5 x – 1) = log 3 (x + 7) log 3 9 = log 3 9 Solution checks. log 3 (5 · 2 – 1) = log 3 (2 + 7) ? Write original equation. Substitute 2 for x.

10. log 5 (3x + 1) = 2 Solving a Logarithmic Equation Solve log 5 (3x + 1) = 2. 3x + 1 = 25 x = 8 The solution is 8. Write original equation. Exponentiate each side Solve for x. Simplify.

11. log 5 (3x + 1) = 2 Solving a Logarithmic Equation Solve log 5 (3x + 1) = 2. 5 = 5 2 log 5 (3x + 1) 3x + 1 = 25 x = 8 The solution is 8. S OLUTION Write original equation. Exponentiate each side using base 5. b = x log b x Solve for x. log 5 (3x + 1) = 2 log 5 (3 · 8 + 1) = 2 ? log 5 25 = 2 ? 2 = 2 Solution checks. C HECK Check the solution by substituting it into the original equation. Simplify. Substitute 8 for x. Write original equation.

12. Because the domain of a logarithmic function generally does not include all real numbers, you should be sure to check for extraneous solutions of logarithmic equations. You can do this algebraically or graphically. Checking for Extraneous Solutions

13. Solve log 5 x + log (x – 1) = 2. Check for extraneous solutions. log [ 5 x (x – 1)] = 2 5 x 2 – 5 x = 100 x 2 – x – 20 = 0 (x – 5 )(x + 4) = 0 x = 5 or x = – 4 S OLUTION log 5 x + log (x – 1) = 2 Write original equation. Product property of logarithms. Exponentiate both sides. Write in standard form. Factor. Zero product property Checking for Extraneous Solutions Simplify

14. The solutions appear to be 5 and – 4. However, when you check these in the original equation you can see that x = 5 is the only solution. S OLUTION log 5 x + log (x – 1) = 2 Check for extraneous solutions. x = 5 or x = – 4 Zero product property Checking for Extraneous Solutions Check: log 5(5) + log(5 – 1) = 2 log 5(-4) + log(-4 – 1) = 2 log 25 + log 4 = 2 log -20 + log -5 = 2 log (25)(4) = 2 error log 100 = 2 2 = 2

15. If necessary use the properties of logarithms to condense several terms into one. If necessary use the properties of logarithms to condense several terms into one. S OLVING L OGARITHMIC E QUATIONS Make sure that there is one term on each side of equation. Always use the inverse property.