Math Chapter 3 Midterm Review Solution Key

Chapter 3: Exponential and Logarithmic Functions 1. Evaluate the following exponential expressions: a. (3^(-5) * 3^(10)) / (3^2) * (3^5 / 3^2) = 3^3 = 27 b. ((4^(-2) + 4^0)^(-1)) / ((1/2 + 1)^(-1)) = 16/17 c. √((9^8 * 3^(-2)) / 3^(-2)) = √(9^8) = 3^4 = 81 d. (2^(6) / 2^(0)) * ((2^(-4) + 2^(-6)) / 2^(-3)) = 5/8 2. Simplify the following expressions: a. 3^(-2) * 3^(5) = 3^(3) = 81 b. √(8^6 / 7^(-1)) = 8^(6/2) * 7^(1/2) = 64√7 c. (6x^(-4/3) + 2x^(5/3)) / (2x^(-2/3)) = 3 + x^3 d. (6x^(1/2) * 2x^(5/2)) / (x^(3/2)) = 3 + x^2 3. Solve each exponential equation: a. 3^(8-x) = 27^(2-x) 3^(8-x) = 3^(3(2-x)) 8-x = 6-3x x = -1 b. 4√(3) * 2^(5/x) = 2^(3x) 2^(2) * 2^(5/x) = 2^(3x) 2^(2 + 5/x) = 2^(3x) 2 + 5/x = 3x x = 9/2 c. 2^(x+1) = 1/16 2^(x+1) = 2^(-4) x + 1 = -4 x = -5 d. √((9^(x+6)) / 27^x) = 3^4 (3^(2(x+6))) / (3^(3x)) = 3^8 3^(-x+6) = 3^8 -x + 6 = 8 x = -2 4. Solve the logarithmic equation. a. log₂x = -4 x = 2^(-4) x = 1/16 b. log₄(x²+5) = log₄(4x²-2x) x²+5 = 4x²-2x 0 = 3x²-2x-5 0 = (3x-5)(x+1) x = 5/3, -1 c. log₂5(25√5) = x 25√5 = 5^x 5² * 5^(1/2) = 5^x 5^(5/2) = 5^x x = 5/2 d. log₆4 + log₆3 - log₆2 = x log₆((4*3)/2) = x log₆6 = x x = 1 e. log₁(x²-4) = 2 x²-4 = 10⁴ x² = 104 x = ±√104 f. log(7-x) - log(3x+2) = 1 log((7-x)/(3x+2)) = 1 (7-x)/(3x+2) = 10¹ 7-x = 30x+20 -13 = 31x x = -13/31 g. log₂(x-2) + log₂(x+5) = 1 log₂((x-2)(x+5)) = 1 x²+3x-10 = 2¹ x²+3x-12 = 0 x = (-3±√57)/2 x = (-3-√57)/2 h. log₄(x-1) = log₄(x-2) - log₄(x+2) log₄(x-1) = log₄((x-2)/(x+2)) x-1 = (x-2)/(x+2) x² = 0 x = 0 5. Solve for x: a. 3^x = 31 x log₃ = log₃31 x = log₃31 b. e^(5x) = 2^(x-3) 5x = (x-3)ln2 5x = 0.693x - 2.079 4.307x = -2.079 x ≈ -0.483 c. e^(2x) - 3e^x + 2 = 0 (e^x - 1)(e^x - 2) = 0 e^x = 1 or e^x = 2 x = 0 or x = ln2 d. 2e^(x-1) - 1 = 4 2e^(x-1) = 5 e^(x-1) = 2.5 x-1 = ln2.5 x = 3 + ln2.5 e. 2^x + 13 = 35 2^x = 22 x = log₂22 f. ln√(x+1) = 2 e² = √(x+1) 4 = x+1 x = 3 g. lnx - ln5 = 4 ln(x/5) = 4 e^4 = x/5 5e^4 = x x = 272.991 h. log₄64 = 3 6. Find the exact value without a calculator: a. log₄64 = 3 b. 27^x = 3 (3³)^x = 3 3^(3x) = 3 3x = 1 x = 1/3 c. 5^(log₄64) = 5³ = 125 7. Given log40 ≈ 1.6021, using rules of logarithms, find the value of the following: a. log₄1000 log₄1000 = log₄(40³)= 3log₄40 = 3 * 1.6021 = 4.8063 b. log₂5 log₂5 = log₂(10/2) = log₂10 - log₂2 = 1.6021 - 1 = 0.6021 8. Express the following in terms of logₐM and logₐN: a. logₐ(M/N) logₐ(M/N) = logₐM - logₐN b. logₐ(M³N²) logₐ(M³N²) = 3logₐM + 2logₐN 9. Expand: a. ln(5e⁻²) ln(5e⁻²) = ln5 + ln(e⁻²) = ln5 - 2 b. ln(x³/(x+3)(xy)) ln(x³/(x+3)(xy)) = ln(x³) - ln(x+3) - ln(x) - ln(y) = 3ln(x) - ln(x+3) - ln(x) - ln(y) 10. Condense: a. ln3 + 1/3 ln(4-x²) - ln(x) ln3 + 1/3 ln(4-x²) - ln(x) = ln(3 * (4-x²)^(1/3) / x) = ln(3√(4-x²)/x) b. 5 ln(x-2) - ln(x+2) - 3 ln(x) 5 ln(x-2) - ln(x+2) - 3 ln(x) = ln((x-2)⁵ / (x+2)x³) 11. A gallon of gasoline cost $3.09 two years ago. Now it costs $1.79. To the nearest percent, what has been the annual rate of decrease in cost? √(1.79/3.09) = 1-r r = 1 - √(1.79/3.09) ≈ 0.241 or 24.1% 12. Suppose $1,200 is invested at an interest rate of 9.6%. How much is the investment worth after 2 years if the interest is compounded as follows:   a. monthly 1200(1 + 0.096/12)^(12*2) ≈ $1452.89 b. continuously 1200e^(0.096*2) ≈ $1454.00 13. If 250 mg of ibuprofen has a half-life of 3 hours, then how much ibuprofen is in a person's bloodstream after the following: a. 7 hours   250(1/2)^(7/3) ≈ 49.61 mg b. t hours 250(1/2)^(t/3) 14. Graph the following and tell how the graphs are related: a. f(x) = 3^x b. f(x) = (1/3)^x c. f(x) = log₃x