For b > 0 and b ≠ 1, if b x = b y, then x = y. S OLVING E XPONENTIAL E QUATIONS If two powers with the same base are equal, then their exponents must be equal.

Solving by Equating Exponents Solve 4 3x = 8 x + 1. ( 2 2 ) 3x = ( 2 3 ) x + 1 6x = 3x x = 2 3x (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 = = 64 Solution checks. Solve for x.

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.

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

x y C HECK Taking a Logarithm of Each Side Check the solution algebraically by substituting into the original equation. Or, check it graphically by graphing both sides of the equation and observing that the two graphs intersect at x ≈ y = 21 y = 10 2 x – Solve 10 2 x – = 21.

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.

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 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.

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.

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

Solve log 5 x + log (x – 1) = 2. Check for extraneous solutions. log [ 5 x (x – 1)] = 2 10 = 10 2 log (5 x 2 – 5 x) 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 each side using base log x = x Write in standard form. Factor. Zero product property Checking for Extraneous Solutions

y x The solutions appear to be 5 and – 4. However, when you check these in the original equation or use a graphic check as shown below, 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 y = 2 y = log 5 x + log (x – 1)

SEISMOLOGY On May 22, 1960, a powerful earthquake took place in Chile. It had a moment magnitude of 9.5. How much energy did this earthquake release? The moment magnitude M of an earthquake that releases energy E (in ergs) can be modeled by this equation: M = ln E Using a Logarithmic Model

M = ln E Using a Logarithmic Model 9.5 = ln E = ln E ≈ ln E e ≈ e ln E x ≈ E The earthquake released about 2.7 trillion ergs of energy. S OLUTION Write model for moment magnitude. Substitute 9.5 for M. Subtract 1.17 from each side. Divide each side by Exponentiate each side using base e. e ln x = e log e x = x

EXPONENTIAL AND LOGARITHMIC PROPERTIES S OLVING L OGARITHMIC E QUATIONS For b > 0 and b ≠ 1, if b x = b y, then x = y. For positive numbers b, x, and y where b ≠ 1, log b x = log b y if and only if x = y. For b > 0 and b ≠ 1, if x = y, then b x = b y.