Copyright © Cengage Learning. All rights reserved. 4 Exponential and Logarithmic Functions Copyright © Cengage Learning. All rights reserved.

4.5 Exponential and Logarithmic Equations Copyright © Cengage Learning. All rights reserved.

Objectives Exponential Equations Logarithmic Equations Compound Interest

Exponential Equations

Exponential Equations An exponential equation is one in which the variable occurs in the exponent. Some exponential equations can be solved by using the fact that exponential functions are one-to-one. This means that ax = ay x = y

Exponential Equations The following guidelines describe the process for solving exponential equations.

Example 2 – Solving an Exponential Equation Consider the exponential equation 3x + 2 = 7. (a) Find the exact solution of the equation expressed in terms of logarithms. (b) Use a calculator to find an approximation to the solution rounded to six decimal places. Solution: (a) We take the common logarithm of each side and use Law 3. 3x + 2 = 7 log(3x + 2) = log 7 Given Equation Take log of each side

Example 2 – Solution (x + 2)log 3 = log 7 x + 2 = cont’d (x + 2)log 3 = log 7 x + 2 = The exact solution is (b) Using a calculator, we find the decimal approximation x  –0.228756. Law 3 (bring down exponent) Divide by log 3 Subtract 2

Example 2 – Solution Check Your Answer cont’d Check Your Answer Substituting x = –0.228756 into the original equation and using a calculator, we get 3(–0.228756) + 2  7

Example 5 – An Exponential Equation of Quadratic Type Solve the equation e2x – ex – 6 = 0. Solution: To isolate the exponential term, we factor. e2x – ex – 6 = 0 (ex)2 – ex – 6 = 0 Given Equation Law of Exponents

Example 5 – Solution (ex – 3)(ex + 2) = 0 ex – 3 = 0 or ex + 2 = 0 cont’d (ex – 3)(ex + 2) = 0 ex – 3 = 0 or ex + 2 = 0 ex = 3 ex = –2 The equation ex = 3 leads to x = ln 3. But the equation ex = –2 has no solution because ex > 0 for all x. Thus x = ln 3  1.0986 is the only solution. Factor (a quadratic in ex) Zero-Product Property

Logarithmic Equations

Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. Some logarithmic equations can be solved by using the fact that logarithmic functions are one-to-one. This means that loga x = loga y x = y

Logarithmic Equations The following guidelines describe the process for solving logarithmic equations.

Example 8 – Solving Logarithmic Equations Solve each equation for x. (a) ln x = 8 (b) log2(25 – x) = 3 Solution: (a) ln x = 8 x = e8 Therefore, x = e8  2981. Given equation Exponential form

Example 8 – Solution We can also solve this problem another way. cont’d We can also solve this problem another way. ln x = 8 eln x = e8 x = e8 Given equation Raise e to each side Property of ln

Example 8 – Solution cont’d (b) The first step is to rewrite the equation in exponential form. log2(25 – x) = 3 25 – x = 23 25 – x = 8 x = 25 – 8 = 17 Given equation Exponential form (or raise 2 to each side)

Example 8 – Solution Check Your Answer If x = 17, we get cont’d Check Your Answer If x = 17, we get log2(25 – 17) = log2 8 = 3

Logarithmic Equations Logarithmic equations are used in determining the amount of light that reaches various depths in a lake. (This information helps biologists to determine the types of life a lake can support.) As light passes through water (or other transparent materials such as glass or plastic), some of the light is absorbed. It’s easy to see that the murkier the water, the more light is absorbed. The exact relationship between light absorption and the distance light travels in a material is described in the next example.

Example 12 – Transparency of a Lake If I0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer-Lambert Law, where k is a constant depending on the type of material.

Example 12 – Transparency of a Lake cont’d (a) Solve the equation for I. (b) For a certain lake k = 0.025, and the light intensity is I0 = 14 lumens (lm). Find the light intensity at a depth of 20 ft.

Example 12(a) – Solution We first isolate the logarithmic term. I = I0e–kx Given equation Multiply by –k Exponential form Multiply by I0

Example 12(b) – Solution We find I using the formula from part (a). cont’d We find I using the formula from part (a). I = I0e–kx = 14e(–0.025)(20)  8.49 The light intensity at a depth of 20 ft is about 8.5 lm. From part (a) I0 = 14, k = 0.025, x = 20 Calculator

Compound Interest

Compound Interest If a principal P is invested at an interest rate r for a period of t years, then the amount A of the investment is given by A = P(1 + r) A(t) = Pert We can use logarithms to determine the time it takes for the principal to increase to a given amount. Simple interest (for one year) Interest compounded n times per year Interest compounded continuously

Example 13 – Finding the Term for an Investment to Double A sum of $5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following methods. (a) Semiannually (b) Continuously

Example 13(a) – Solution We use the formula for compound interest with P = $5000, A(t) = $10,000, r = 0.05, and n = 2, and solve the resulting exponential equation for t. (1.025)2t = 2 log 1.0252t = log 2 2t log 1.025 = log 2 Divide by 5000 Take log of each side Law 3 (bring down the exponent)

Example 13(a) – Solution t  14.04 cont’d t  14.04 The money will double in 14.04 years. Divide by 2 log 1.025 Calculator

Example 13(b) – Solution cont’d We use the formula for continuously compounded interest with P = $5000, A(t) = $10,000, and r = 0.05 and solve the resulting exponential equation for t. 5000e0.05t = 10,000 e0.05t = 2 ln e0.05t = ln 2 0.05t = ln 2 Pert = A Divide by 5000 Take ln of each side Property of ln

Example 13(b) – Solution t  13.86 cont’d t  13.86 The money will double in 13.86 years. Divide by 0.05 Calculator