Lesson 9 – 5 Exponential Equations & Inequalities Pre-calculus

Learning Objective To solve exponential equations To solve continuous interest problems

Logarithmic Vocabulary Consider: log 260 = log (2.6 𝑥 10 2 ) = log 2.6 + log 10 2 =0.4150+2 =2.4150 mantissa characteristic Also log 0.26 = log (2.6 𝑥 10 −1 ) = log 2.6 + log −1 =0.4150+(−1) =−0.5850 mantissa characteristic

Logarithmic Vocabulary 1. Underline the mantissa & circle the characteristic Logarithmic Vocabulary log 425 =2.6284 If we are given log 𝑥 or ln 𝑥 , we can find 𝑥 using our calculators. We will use 10 𝑥 or 𝑒 𝑥 Let’s practice some simple ones…get those calculators ready!

Logarithmic Equations Solve for 𝑥 to the nearest hundredth. Logarithmic Equations 2. log 𝑥 =3.2274 10 3.2274 =𝑥 𝑥=1688.11 Around the world! 3. 2log 𝑥 =2.6419 log 𝑥 =1.32095 10 1.32095 =𝑥 𝑥=20.94 4. ln 𝑥 −3=5.7213 ln 𝑥 =8.7213 𝑒 8.7213 =𝑥 𝑥=6132.15

Check – up Solve for 𝑥 to the nearest hundredth. 1. log 𝑥 =2.3010 𝑥=199.99

Exponential Equations To solve an exponential equation: Exponential Equations 1. Isolate the exponential expression 2. Take the logarithm of both sides of the equation 3. Verify all answers! (by substitution in original)

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 5. 4 2𝑥−1 −27=0 4 2𝑥−1 =27 log 4 2𝑥−1 = log 27 (2𝑥−1) log 4 = log 27 2𝑥 log 4 − log 4 = log 27 2𝑥 log 4 = log 27 + log 4 𝑥= log 27 + log 4 2 log 4 𝑥=1.69

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 6. 3 16 −2𝑥+1 −0.71=0 3 16 −2𝑥+1 =0.71 16 −2𝑥+1 = 0.71 3 Move decimal 2 places log 16 −2𝑥+1 = log 71 300 (−2𝑥+1) log 16 = log 71 300 Distribute −2𝑥 log 16 + log 16 = log 71 300 −2𝑥 log 16 = log 71 300 − log 16 𝑥= log 71 300 − log 16 −2 log 16 𝑥=0.76

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 7. 3 7 4𝑥−1 =4( 2 7𝑥+3 ) log( 3 7 4𝑥−1 ) = log (4 2 7𝑥+3 ) Expand logs log 3 + log 7 4𝑥−1 = log 4 + log ( 2 7𝑥+3 ) log 3 + 4𝑥−1 log 7 = log 4 +(7𝑥+3) log 2 log 3 +4𝑥 log 7 − log 7 = log 4 +7𝑥 log 2 +3 log 2 4𝑥 log 7 −7𝑥 log 2 = log 4 +3 log 2 − log 3 + log 7 𝑥(4 log 7 −7 log 2 )= log 4 +3 log 2 − log 3 + log 7 𝑥= log 4 +3 log 2 − log 3 + log 7 4 log 7 −7 log 2 𝑥=1.47

Solve for 𝑥 to nearest hundredth. Check – up 2. 4 2𝑥 = 6 𝑥+1 𝑥=1.83

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 8. 𝑒 𝑥 𝑒 𝑥 −1 =5 𝑒 𝑥 =5( 𝑒 𝑥 −1) 𝑒 𝑥 =5 𝑒 𝑥 −5 −4 𝑒 𝑥 =−5 𝑒 𝑥 = 5 4 ln 𝑒 𝑥 = ln 5 4 𝑥= ln 5 4 𝑥=0.22

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 9. 2 𝑒 𝑥 −5=3 𝑒 −𝑥 ( ) 𝑒 𝑥 2 𝑒 𝑥 −5−3 𝑒 −𝑥 =0 2 𝑒 2𝑥 −5 𝑒 𝑥 −3=0 2 𝑒 𝑥 +1 𝑒 𝑥 −3 =0 𝑒 𝑥 =− 1 2 𝑒 𝑥 =3 𝑥= ln − 1 2 𝑥= ln 3 𝑥=1.10

Exponential Equations Solve for 𝑥 to nearest hundredth. Exponential Equations 10. 8 𝑒 𝑥 +5 𝑒 −𝑥 −14=0 ( ) 𝑒 𝑥 8 𝑒 2𝑥 −14 𝑒 𝑥 +5=0 4 𝑒 𝑥 −5 2 𝑒 𝑥 −1 =0 𝑒 𝑥 = 5 4 𝑒 𝑥 = 1 2 𝑥= ln 5 4 𝑥= ln 1 2 𝑥=0.22 𝑥=−0.69

Solve for 𝑥 to nearest hundredth. Check – up 3. 4 𝑥 4 𝑥 −5 =3 𝑥=1.45

Exponential Equations Sometimes we can’t solve algebraically, so we go to our graphing calculator. Exponential Equations Solve using a graphing calculator. 11. 𝑒 𝑥 = 𝑥 2 −1 𝑌 1 = 𝑒 𝑥 (Find intersection) 𝑌 2 = 𝑥 2 −1 𝑥=−1.15 12. 𝑦≥ 𝑒 𝑥 −2 𝑌 1 = 𝑒 𝑥 −2

Compound Interest Applications Compound Interest Formula: 𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡 𝐴 = total value of investment 𝑡 = number of years 𝑃 = principal amount invested 𝑟 = interest rate (%  decimal) 𝑛 = number of times per year interest is compounded

Compound Interest 𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡 𝐴 = 20,000 𝑟 = 0.083 𝑛 = 12 𝑃 = 11,500 13. The Smith Family wants to give their youngest daughter $20,000 when she is ready for college. They now have $11,500 to invest. Determine how many years it will take them to achieve their goal given that they invest this amount at 8.3% compounded monthly. Compound Interest 𝐴=𝑃 1+ 𝑟 𝑛 𝑛𝑡 𝐴 = 20,000 𝑟 = 0.083 𝑃 = 11,500 𝑛 = 12 𝑡 = ? 20,000=11,500 1+ 0.083 12 12𝑡 20,000 11,500 = 1+ 0.083 12 12𝑡 log log log 20,000 − log 11,500 =12𝑡 log 1+ 0.083 12 𝑡= log 20,000 − log 11,500 12 log 1+ 0.083 12 𝑡=7 𝑦𝑒𝑎𝑟𝑠 *Watch those parentheses!

Compound Interest Continuous Compound Interest Formula 𝐴=𝑃 𝑒 𝑟𝑡 14. A sum of money invested at a fixed interest rate, compounded continuously, tripled in 19 years. Determine the interest rate at which the money was invested. You don’t know A or P but you don’t need it! You need P to triple 𝐴=𝑃 𝑒 𝑟𝑡 ln 3 =19𝑟 ln 𝑒 3𝑃=𝑃 𝑒 𝑟𝑡 ln 3 19 =𝑟 3= 𝑒 𝑟𝑡 𝑟=5.8%

4. You invested $4200 in an account paying 6 4. You invested $4200 in an account paying 6.5% compounded continuously. How long to the nearest year will it take for the money in the account to increase by $900? Check – up 3 years

Assignment Pg. 472 #1, 3, 5, 9, 13, 18, 20–23, 25, 27, 29, 32, 33, 36, 38, 39–47 odd