Math Chapter 4 Review Solution Key

Chapter 4 Review Graph the function. Identify the domain and range in interval notation and the horizontal or vertical asymptote. y = -2^x + 5   Domain: (-∞, ∞) Range: (-∞, 5) HA: y = 5 y = 2^(x-1) - 3 Domain: (-∞, ∞) Range: (-3, ∞) HA: y = -3 y = log₃(x-2) Domain: (2, ∞) Range: (-∞, ∞) VA: x = 2 y = 2 - log₃(x+1) Domain: (-1, ∞) Range: (-∞, ∞) VA: x = -1 In #5, find a function of the form y = a^x for the given graph. In #6, find a function of the form y = logₐx for the given graph. y = 1/4^x y = log₂x 7. Suppose $1000 is invested in an account that earns 10% annual interest. Find the amount of money earned after 5 years if it is compounded a) Semiannually 1000(1   + 10/2)^(2*5) ≈ $1628.89 b) Continuously 1000e^(0.10*5) ≈ $1648.72 Express the function in its opposite form 3^x = 81 log₃81 = x e^x = 2 ln2 = x log₄16 = 2 4² = 16 ln x = 5 e⁵ = x Evaluate the expression log₇343 7^x = 343 x = 3 log₆√6 6^x = 6^(1/2) x = 1/2 log₅25 log₅(5²) = 2log₅5 = 2 e^ln6 6 log₂4 2^x = 4 x = 2 log₉(1/81) 9^x = 1/81 9^x = 9^(-2) x = -2 log₉√3 9^x = 3^(1/2) 3^(2x) = 3^(1/2) 2x = 1/2 x = 1/4 log₄2 + log₄128 log₄(2 * 128) = log₄256 = x 4^x = 256 x = 4 log₃189 - log₃7 log₃(189/7) = log₃27 = x 3^x = 27 x = 3 Find the domain of the function y = log₂(x + 3) x + 3 > 0 x > -3 (-3, ∞) y = ln(2 - 5x) 2 - 5x > 0 -5x > -2 x < 2/5 (-∞, 2/5) y = log₅x + 2 x > 0 (0, ∞) Use the Laws of Logarithms to condense the expression ln 6 + 5 ln x + 9 ln(x² + 9) ln(6x⁵(x²+9)⁹) 2 log x - 4 log y + (1/2) log z log(x²/y⁴√z) Use the Laws of Logarithms to expand the expression log(x²/yz²) log x² - log y - log z² 2 log x - log y - 2 log z log√((x-1)/x²) (1/2)(log(x-1) - 2log x) Solve the EXPONENTIAL equation 16^x = 32 2^(4x) = 2^5 4x = 5 x = 5/4 3^(x-4) = 1/9 3^(x-4) = 3^(-2) x - 4 = -2 x = 2 (1/64)^x = 256^(x+2) 4^(-3x) = 4^(4(x+2)) -3x = 4x + 8 -7x = 8 x = -8/7 9^(2x) = 3^(x+4) 3^(4x) = 3^(x+4) 4x = x + 4 3x = 4 x = 4/3 2^(2x-2) = 2^(x+9) 2x - 2 = x + 9 x = 11 2^(2x) = 3 2x = log₂3 x = (1/2)log₂3 Solve the EXPONENTIAL equation. Round answers to the nearest thousandth. 3^x = 26 x log 3 = log 26 x = log 26 / log 3 ≈ 2.967 e^(2-3x) = 12 2 - 3x = ln 12 -3x = -2 + ln 12 x = (-2 + ln 12) / -3 ≈ 1 12^x = 5^(x+4) x log 12 = (x+4) log 5 1.079x = 0.699x + 2.796 0.38x = 2.796 x ≈ 7.36 x^3 * 9^x - 9^x = 0 9^x(x³ - 1) = 0 9^x = 0 (no solution) x³ - 1 = 0 x³ = 1 x = 1 e^(2x) - 5e^x + 4 = 0 (e^x - 4)(e^x - 1) = 0 e^x = 4 or e^x = 1 ln 4 = x or ln 1 = x 1.386, 0 e^x * x² + 4e^x * x = 21e^x e^x(x² + 4x - 21) = 0 e^x(x+7)(x-3) = 0 x = -7, 3 Solve the LOGARITHMIC equation. Round answers to the nearest thousandth. ln x = 5 x = e^5 ≈ 148.413 5 - 2log(9x+6) = 1 -2log(9x+6) = -4 log(9x+6) = 2 100 = 9x + 6 x = 10.444 8 + 3log₃(4-x) = 17 3log₃(4-x) = 9 log₃(4-x) = 3 3³ = 4 - x 27 = 4 - x x = -23 log₃(x+4) - log₃(x-4) = 3 log₃((x+4)/(x-4)) = 3 (x+4)/(x-4) = 27 x + 4 = 27(x-4) x + 4 = 27x - 108 26x = 112 x = 4.308 log(x+2) + log(x-1) = 1 log((x+2)(x-1)) = 1 (x+2)(x-1) = 10 x² + x - 12 = 0 (x+4)(x-3) = 0 x = -4, 3 log₈(x²-x) = log₈12 x² - x = 12 x² - x - 12 = 0 (x-4)(x+3) = 0 x = 4, -3 Find the time required for an investment of $3000 to increase to $8000 if it is compounded quarterly at 8% annual interest.   8000 = 3000(1 + 0.08/4)^(4t) (8/3) = (1.02)^(4t) log(8/3) = 4t log 1.02 t = log(8/3) / (4 log 1.02) ≈ 12.38 years Find the time it would take for an investment to triple in value if it is compounded quarterly at an interest rate of 1.25%.   3 = 1(1 + 0.0125/4)^(4t) 3 = 1.003125^(4t) log 3 = 4t log 1.003125 t = log 3 / (4 log 1.003125) ≈ 88.03 years 48. The number of bacteria in a culture is given by n(t) = 400e^(0.49t), where t is measured in hours. a) How many bacteria remain after 5 hours? n(5) = 400e^(0.49*5) ≈ 4635.34 bacteria b) After how many hours will the number of bacteria reach 10,000? 10000 = 400e^(0.49t) 25 = e^(0.49t) ln 25 = 0.49t t = ln(25) / 0.49 ≈ 6.57 hours 49. A bacteria culture contains 500 bacteria initially and doubles every 15 minutes. a) Find a function that models the number of bacteria at time t.   1000 = 500e^(15r) 2 = e^(15r) ln 2 = 15r r = ln(2)/15 ≈ 0.0462 n(t) = 500e^(0.0462t) b) Find the number of bacteria after 3 hours. n(180) = 500e^(0.0462*180) ≈ 2,044,386.08 c) After how many minutes will there be 5000 bacteria? 5000 = 500e^(0.0462*t) 10 = e^(0.0462*t) ln 10 = 0.0462t t = ln(10) / 0.0462 ≈ 49.84 minutes 50. The fox population in a certain region has a growth rate of 6% per year. It is estimated that the population in 2000 was 16,000 foxes. a) Find a function that models the population t years after 2000. n(t)   = 16000e^(0.06t) b) Estimate the fox population in 2012. n(12) = 16000e^(0.06*12) ≈ 32870.93 51. Ten years ago, Michael paid $250 for a rare 1923 stamp. Its current value is $1,000. Find the average annual rate of growth. 1000 = 250(1+r)^10 4 = (1+r)^10 r = ∛4 - 1 ≈ 14.87% 52. A baseball card increased in value from $50 to $500 in 15 years. Find its average annual rate of appreciation. 500 = 50(1+r)^15 10 = (1+r)^15 ∛10 = 1+r r = ∛10 - 1 ≈ 16.59%