Exponential and Logarithmic Functions Copyright Cengage Learning All

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

Exponential and 4. 5 Logarithmic Equations Copyright © Cengage Learning. All rights reserved.

Objectives ► Exponential Equations ► Logarithmic Equations ► Compound Interest 3

Exponential Equations 4

Exponential Equations An exponential equation is one in which the variable occurs in the exponent. For example, 2 x = 7 The variable x presents a difficulty because it is in the exponent. To deal with this difficulty, we take the logarithm of each side and then use the Laws of Logarithms to “bring down x” from the exponent. 5

Exponential Equations 2 x = 7 ln 2 x = ln 7 x ln 2 = ln 7 Given Equation Take In of each side Law 3 (Bring down exponent) Solve for x 2. 807 Calculator 6

Solving Exponential Equations Find the solution of the equation 3 x + 2 = 7, rounded to six decimal places 7

Example 1 – Solving an Exponential Equation Find the solution of the equation 3 x + 2 = 7, rounded to six decimal places. Solution: We take the common logarithm of each side and use Law 3. 3 x + 2 = 7 log(3 x + 2) = log 7 Given Equation Take log of each side 8

Example 1 – Solution (x + 2)log 3 = log 7 x+2= cont’d Law 3 (bring down exponent) Divide by log 3 Subtract 2 – 0. 228756 Calculator 9

Example 1 – Solution cont’d Check Your Answer Substituting x = – 0. 228756 into the original equation and using a calculator, we get 3(– 0. 228756) + 2 7 10

Example 4 – An Exponential Equation of Quadratic Type Solve the equation e 2 x – ex – 6 = 0. Solution: To isolate the exponential term, we factor. e 2 x – ex – 6 = 0 (ex)2 – ex – 6 = 0 Given Equation Law of Exponents 11

Example 4 – Solution (ex – 3)(ex + 2) = 0 ex – 3 = 0 ex = 3 or cont’d Factor (a quadratic in ex) ex + 2 = 0 Zero-Product Property 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. 12

Logarithmic Equations 13

Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. For example, log 2(x + 2) = 5 To solve for x, we write the equation in exponential form. x + 2 = 25 x = 32 – 2 = 30 Exponential form Solve for x 14

Logarithmic Equations The method used to solve this simple problem is typical. We summarize the steps as follows. Solve each equation for x. (a) ln x = 8 (b) log 2(25 – x) = 3 15

Example 6 – Solving Logarithmic Equations Solve each equation for x. (a) ln x = 8 (b) log 2(25 – x) = 3 Solution: (a) ln x = 8 x = e 8 Given equation Exponential form Therefore, x = e 8 2981. 16

Example 6 – Solution cont’d We can also solve this problem another way: ln x = 8 eln x = e 8 Given equation Raise e to each side Property of ln 17

Example 6 – Solution cont’d (b) The first step is to rewrite the equation in exponential form. log 2(25 – x) = 3 25 – x = 23 Given equation Exponential form (or raise 2 to each side) 25 – x = 8 x = 25 – 8 = 17 18

Example 6 – Solution cont’d Check Your Answer If x = 17, we get log 2(25 – 17) = log 2 8 = 3 19

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. 20

Example 10 – Transparency of a Lake If I 0 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. 21

Example 10 – 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 I 0 = 14 lumens (lm). Find the light intensity at a depth of 20 ft. 22

Example 10 – Solution (a) We first isolate the logarithmic term. Given equation Multiply by –k Exponential form I = I 0 e–kx Multiply by I 0 23

Example 10 – Solution cont’d (b) We find I using the formula from part (a). I = I 0 e–kx From part (a) = 14 e(– 0. 025)(20) I 0 = 14, k = 0. 025, x = 20 8. 49 Calculator The light intensity at a depth of 20 ft is about 8. 5 lm. 24

Compound Interest 25

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) Simple interest (for one year) Interest compounded n times per year A(t) = Pert Interest compounded continuously We can use logarithms to determine the time it takes for the principal to increase to a given amount. 26

Example 11 – 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 method. (a) Semiannually (b) Continuously 27

Example 11 – Solution (a) 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)2 t = 2 log 1. 0252 t = log 2 2 t log 1. 025 = log 2 Divide by 5000 Take log of each side Law 3 (bring down the exponent) 28

Example 11 – Solution cont’d Divide by 2 log 1. 025 t 14. 04 Calculator The money will double in 14. 04 years. 29

Example 11 – Solution cont’d (b) 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. 5000 e 0. 05 t = 10, 000 e 0. 05 t = 2 ln e 0. 05 t = ln 2 Pert = A Divide by 5000 Take ln of each side Property of ln 30

Example 11 – Solution cont’d Divide by 0. 05 t 13. 86 Calculator The money will double in 13. 86 years. 31