5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS Copyright © Cengage Learning. All rights reserved.

Solve simple exponential and logarithmic equations. Solve more complicated exponential equations. Solve more complicated logarithmic equations. Use exponential and logarithmic equations to model and solve real-life problems. What You Should Learn

Introduction There are two basic strategies for solving exponential or logarithmic equations. The first is based on the One-to-One Properties and was used to solve simple exponential and logarithmic equations. The second is based on the Inverse Properties. For a > 0 and a ≠ 1, the following properties are true for all x and y for which loga x and loga y are defined.

Introduction One-to-One Properties ax = ay if and only if x = y. loga x = loga y if and only if x = y. Inverse Properties aloga x = x loga ax = x

Example 1 – Solving Simple Equations Original Rewritten Equation Equation Solution Property a. 2x = 32 2x = 25 x = 5 One-to-One b. ln x – ln 3 = 0 ln x = ln 3 x = 3 One-to-One c. = 9 3– x = 32 x = –2 One-to-One d. ex = 7 ln ex = ln 7 x = ln 7 Inverse e. ln x = –3 eln x = e– 3 x = e–3 Inverse f. log x = –1 10log x = 10–1 x = 10–1 = Inverse g. log3 x = 4 3log3 x = 34 x = 81 Inverse

Introduction The strategies used in Example 1 are summarized as follows. You may use whichever one of the 3 strategies you think is appropriate for the problem you are solving:

Solving Exponential Equations

Example 2 – Solving Exponential Equations Solve each equation and approximate the result to three decimal places, if necessary. a. e – x2 = e – 3x – 4 b. 3(2x) = 42

Example 2(a) – Solution e –x2 = e – 3x – 4 –x2 = –3x – 4 x2 – 3x – 4 = 0 (x + 1)(x – 4) = 0 (x + 1) = 0 (x – 4) = 0 The solutions are x = –1 and x = 4. Check these in the original equation. Set 1st factor equal to 0. Set 2nd factor equal to 0. Write original equation. One-to-One Property Write in general form. Factor. x = 4 x = –1

Example 2(b) – Solution 3(2x) = 42 2x = 14 log2 2x = log2 14 x = log2 14 x =  3.807 The solution is x = log2 14  3.807. Check this in the original equation. cont’d Write original equation. Divide each side by 3. Take log (base 2) of each side. Inverse Property Change-of-base formula

Example 4 – Solving an Exponential Equation Solve: 2 3 2𝑡−5 −4=11 Add 4 to both sides: 2 3 2𝑡−5 =15 Divide both sides by 2: 3 2𝑡−5 = 15 2 Take log base 3 of both sides: log 3 3 2𝑡−5 = log 3 7.5 Use the inverse property: 2𝑡−5= log 3 7.5 Solve for t: 𝑡= log 3 7.5 + 5 2 ≈ 1.834 + 5 2 ≈3.417

Example 5 – Exponential Equation of Quadratic Type Solve: 𝑒 2𝑥 −3 𝑒 𝑥 +2=0 Write in quadratic form: 𝑒 𝑥 2 −3 𝑒 𝑥 +2=0 Factor: (𝑒 𝑥 −2)( 𝑒 𝑥 −1)=0 Set each factor to 0: 𝑒 𝑥 −2=0 𝑒 𝑥 −1=0 𝑒 𝑥 =2 𝑒 𝑥 =1 Take the natural ln 𝑒 𝑥 = ln 2 ln 𝑒 𝑥 = ln 1 log of both sides: Use inverse property: 𝑥= ln 2 𝑥= ln 1=0

Another Example Solve: 400 1+ 𝑒 −𝑥 =350 Multiply both sides by the LCD: 400=350 1+ 𝑒 −𝑥 Divide both sides by 350: 400 350 =1+ 𝑒 −𝑥 Reduce 400/350 and subtract 1 from both sides: 8 7 −1= 𝑒 −𝑥 or 1 7 = e −𝑥 Take natural log of both sides: ln 1 7 = ln 𝑒 −𝑥 Use inverse property: ln 1 7 =−𝑥 or 𝑥=− ln 1 7

Solving Logarithmic Equations

Solving Logarithmic Equations To solve a logarithmic equation, you can write it in exponential form. ln x = 3 eln x = e3 x = e3 This procedure is called exponentiating each side of an equation. Logarithmic form Exponentiate each side. Exponential form

Example 6 – Solving Logarithmic Equations a. ln x = 2 eln x = e2 x = e2 b. log3(5x – 1) = log3(x + 7) 5x – 1 = x + 7 4x = 8 x = 2 Original equation Exponentiate each side. Inverse Property Original equation One-to-One Property Add –x and 1 to each side. Divide each side by 4.

Example 6 – Solving Logarithmic Equations c. log6(3x + 14) – log6 5 = log6 2x 3x + 14 = 10x –7x = –14 x = 2 Quotient Property of Logarithms Original equation One-to-One Property Multiply both sides by 5. Isolate x. Divide each side by –7. cont’d

Example – Solving a Natural Log Equation Solve: 2 ln 3𝑥+5=9 2 ln 3𝑥=4 ln 3𝑥=2 𝑒 ln 3𝑥 = 𝑒 2 3𝑥= 𝑒 2 𝑥= 𝑒 2 3 Subtract 5 from both sides: Use the inverse property: Divide both sides by 2: Divide both sides by 3: Exponentiate both sides:

Another Example Solve: ln 𝑥+1 − ln (𝑥−2) = ln 𝑥 ln 𝑥+1 𝑥−2 = ln 𝑥 𝑥+1 𝑥−2 =𝑥 𝑥+1=𝑥 𝑥−2 𝑥+1= 𝑥 2 −2𝑥 0= 𝑥 2 −3𝑥−1 𝑥= 3± 9−4(1)(−1) 2 = 3± 13 2 Use the Quotient Property of logs: Use the one-to-one property: Put in general form: Multiply both sides by the LCD: Distribute the x: Quadratic Formula:

Note on Extraneous Solutions In the previous example, we had the following equation: ln 𝑥+1 − ln (𝑥−2) = ln 𝑥 And we had the following solutions: 𝑥= 3± 13 2 Remember that you cannot take the log of 0 or a negative number. If we plug 3− 13 2 into the original equation, we will have to take the natural log of negative numbers. So, that solution must be thrown out. Thus, there is only one solution: 𝑥= 3+ 13 2

Applications

Example 10 – Doubling an Investment You have deposited $500 in an account that pays 6.75% interest, compounded continuously. How long will it take your money to double? Solution: Using the formula for continuous compounding, you can find that the balance in the account is A = Pert A = 500e0.0675t.

Example 10 – Solution To find the time required for the balance to double, let A = 1000 and solve the resulting equation for t. 500e0.0675t = 1000 e0.0675t = 2 ln e0.0675t = ln 2 0.0675t = ln 2 t = t  10.27 cont’d Let A = 1000. Take natural log of each side. Divide each side by 500. Inverse Property Divide each side by 0.0675. Use a calculator.

Example 10 – Solution The balance in the account will double after approximately 10.27 years. This result is demonstrated graphically in Figure 5.31. cont’d Figure 5.31