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Name: Date: Using Logarithms to solve Bx = A 1. Given the exponential equation Bx = A, use the change of base formula to solve for x. x = lnA/lnB 2. Given the logarithmic equation; log B A x , use the change of base formula to solve for x. x = lnA/lnB Solve: 3. 25x = 125 x log 125 3 log 25 2 6. 4x = 100 x log 100 3.32 log 4 5. 36x = 216 x 4 12.8 1.891 x 7. 5x = .04 8. 10x = 125 x 9. 2x-2 = 128 x2 4. x4 = 12.8 ln 0.04 2 ln 5 x log 125; x 2.097 10. 8x+3 = 500 log 128 ;x 9 log 2 12. 34x+1 = 85 log 85 4x 1 ; x .761 log 3 x x3 11. 252x = 125 log 500 ; x 0.011 log 8 2 x 40 6 x 1 216 x log 40 ; x 5.322 log 2 x 1.08 x 12.5 1.025 x 8 / 13 .18 18. 1501 12 1.01512 x 37 / 3 12 x 12 x 1850 log 37 / 3 ; x 14.062 log 1.015 log 125 ; x 0.75 log 25 14. 4(6)x-1 = 864 16. 650(1.025) = 400 log 12.5 ; x 32.32 log 1.08 2x 13. 3(2)x = 120 15.b 100(1.08) = 1250 x log 216 3 log 36 2 x log 8 / 13 ; x 19.662 log 1.025 .25 19. 90001 4 .9375 4 x 5 / 6 4x 4x x 1 log 216 ;x 4 log 6 .04 17. 3001 2 1.02 2 x 2.5 2x 2x 750 log 2.5 ; x 23.136 log 1.02 52 x 1.08 7500 20. 35001 20000 52 1327 / 1300 52 x 40 / 7 log 5 / 6 ; x .706 log .9375 52 x log 40 / 7 ; x 1.631 log 1327 / 1300 Properties of Logarithms A. Rewrite as a single logarithm B. Evaluate your answer from part A 21) log x 4 log 2 1 22) log 5 u log 5 8 3 log 16 1.204 log 5 2 0.431 23) log m log 3 log 7 24) log y log 35 log 7 log 21 1.322 log 5 0.699 25) log p log 4 3 log 5 26) 4 log x log 32 log 2 log x 4 log 16 log 4 (log 5 ) 3 x 4 16, x 2 log 500 2.699 log 16 2 27) log 2 x 2 log 6 log 3 28) log 2 t 2 log 2 5 3 log 2 2 3 t log 2 25 log 2 8 3 t log 2 log 2 200 3 t 200; 3 t 600 log 2 log 2 x log 36 log 3 2 x 108; x 54 Write each number as a decimal (do as many as possible without a calculator) 29) ln e14 14 31) 5 ln e 5 32) 5 ln e 2 10 33) log 18 18 20 y 34) 7 log 3 3 8 log 3 3 y 34) log 8 100 log 3 37 log 3 38 log 3 y = 20 37 38 1 8 log 100 log 100 1 1 or, 1 1 1 3 2 8 log 10 log 10 4 7 log 3 3 8log 3 3 7(1) 8(1) 4 log 3 8 Write in exponential form 33) ln 42 3.738 34) ln 2.4 0.875 35) ln e3 y e 3.738 42 e 0.875 2.4 e y e3 Write in logarithmic form 36) e 7 1097 37) e x 5.755 38) 2e x 12.8 ln 1097 7 ln 5.755 x ln 6.4 x or ln12.8 x 2 39) Danny’s college savings are invested in a bond that pays an annual interest rate of 6.2% compounded continuously. How long will it take the money to triple? 3P Pe0.062t 3 e0.062t ln 3 .062t t ln 3 t 17.720 yrs .062 40) At what rate would you have to invest your money for it to double if it were compounded continuously for 8 years? 2 e 8 r ln 2 8r r ln 2 rate 8.7% 8 41) At what rate would you have to invest your money for it to double if it were compounded monthly for 2 years? 24 r r r 2 1 24 2 1 24 2 1 12 24 2 1 35.16% 12 12 12 42) How long will it take $200 to grow to $1000 if it was invested @ an annual rate 8% compounded quarterly? 4t ln 5 1 ln 5 .08 1000 2001 5 1.02 4t 4t t t 20.318 yrs 4 ln 1.02 4 ln 1.02 43) When a certain medical drug is administered to a patient, the number of milligrams remaining in the patient’s bloodstream after t hours is modeled by D(t) = 50e-0.2t A) Create a word model to describe the function D(t) 50 mg are injected and the drug decreases exponentially by 20% every hour B) How many milligrams would remain in the blood after 3 hours? D(t) = 50e-0.2(3) = 27.44 mg C) Due to the toxicity of the drug and the possibility negative side effects, patients should not take another dose until only 10% of the drug remains in the system. Approximately how long after the first dose should you wait until taking a second dose? 5 = 50e-0.2(t) 0.1 = e-0.2(t) ln0.1 = -0.2t t = ln0.1/-.02 -5ln0.1 about 11.51 years 44) Nancy wants to invest $4000 in savings certificates that bear an interest rate of 5.75% compounded semiannually. How long should she invest for if she wants to make $1200 in interest? 2t 2t ln 1.3 0.0575 0.0575 5200 40001 t 4.628 years 1.3 1 2t 2 2 ln 1.08875 45) Radioactive iodine decays exponentially in mass by 8.7% every day. If a sample began with 20 grams, how long would it take until there were 10 grams remaining? 10 = 20e-0.087t 0.5 = e-0.087t t = ln0.5/-0.87: about 7.96 years 46) On January 1st 2002 the population of a country had a population of 4,100,000 people. Since that time it has been growing at an annual rate of 3.5% (this is not continuous/exponential). Assume that this has been and always will remain the same. A) What was the population of the country 6 months later? 4100000(1.035)(0.5): about 4,171,133 B) What was the population of the country 9 years later? 4100000(1.035)(9): about 5,587,879 C) What will be the population of the country two decades later? 4100000(1.035)(20): about 5,8158,134 D) According to this model what would the population have been in 1970? 4100000(1.035)(-32): about 1,363,618 47) Find the missing variable and create a word problem to match each problem. A) 4200 = 9000e15r $9000 decreases exponentially. What is the rate that leaves you with $4200 after 15 years? B) 4200 = 9000(r)15 $9000 decreases annually. What is the rate that leaves you with $4200 after 15 years? r C) 42001 12 180 $4200 is compounded monthly for 15 years at an unknown rate.