Math Chapter 3 Review Solution Key

Here's a breakdown of the information from the image you provided: Chapter 3 Review Simplify the following without using a calculator. log₃1 = 0 log₃(81/49) = log₃81 - log₃49 = 4 - 2 = 2 log₁₀(100x²) = log₁₀100 + log₁₀x² = 2 + 2log₁₀x log₃123 ≈ 13 log₅(log₅5) = log₅1 = 0 log₃(1/81) = log₃3^(-4) = -4 log₆(1/6) = log₆6^(-1) = -1 log₂√32 = log₂(2^(5/2)) = 5/2 -3ln(1/e) = -3(-1) = 3 -2ln(e²) + 1 = -2(2) + 1 = -3 logM + logA - logT - 2logH = log((MA)/(TH²)) -log₇0.75 ≈ 0.15 3log₁₅5 ≈ 4.33 log₃6 ≈ 1.63 Solve the following for x. Round to the nearest hundredth when necessary. 4^(x+2) = 32 2^(2x+4) = 2^5 2x + 4 = 5 x = 1/2 2^(2x+2) = 64 2^(2x+2) = 2^6 2x + 2 = 6 x = 2 1/n^(3x) = 4/n^x 4n^(3x) = n^x 2²n^(3x) = n^x 2 + 3x = x x = -1 log₅(1/25) = x log₅5^(-2) = x -2 = x log₃8 = x x ≈ 1.89 16^x = 8^(2x-1) 2^(4x) = 2^(3(2x-1)) 4x = 6x - 3 x = 3/2 (x-4)² = 16 x-4 = ±4 x = 6, 2 ln(x+1) = 9 x+1 = e^9 x = e^9 - 1 2x + ln(e^(4x)) = 12 2x + 4x = 12 x = 2 10^(2x) - 9*10^x + 14 = 0 (10^x - 2)(10^x - 7) = 0 10^x = 2 or 10^x = 7 x = log₁₀2 or x = log₁₀7 3^(1+4x) - 4*9^x + 1 = 0 3*3^(4x) - 4*(3²)^(x) + 1 = 0 3u² - 4u + 1 = 0 (where u = 3^x) (3u-1)(u-1) = 0 u = 1/3 or u = 1 3^x = 1/3 or 3^x = 1 x = -1 or x = 0 (x+1)^(-5/3) = -32 (x+1)^(-5/3) = (-2)^5 x+1 = (-2)^(-3/5) x+1 = -1/8 x = -9/8 Solve the following for x. Round to the nearest hundredth when necessary. 2^(x+1) = 100 log₂(100) = x + 1 x = log₂(100) - 1 ≈ 5.64 3^(1-x) = 5 log₃(5) = 1 - x x = 1 - log₃(5) ≈ 0.46 logₓ(16) = 5 x⁵ = 16 x = 16^(1/5) ≈ 1.74 log(x) + log(x+3) = 1 log(x(x+3)) = 1 x² + 3x = 10 (x+5)(x-2) = 0 x = -5, 2 log(3x+7) + log(x-2) = 1 log((3x+7)(x-2)) = 1 (3x+7)(x-2) = 10 3x² + x - 24 = 0 (3x-8)(x+3) = 0 x = 8/3, -3 log₄(x²+5) = log₄(4x²-2x) x² + 5 = 4x² - 2x 0 = 3x² - 2x - 5 0 = (3x-5)(x+1) x = 5/3, -1 log(7-x) - log(3x+2) = 1 log((7-x)/(3x+2)) = 1 (7-x)/(3x+2) = 10 7-x = 30x+20 -13 = 31x x = -13/31 ln(7-x) + ln(3x+5) = ln(24x) ln((7-x)(3x+5)) = ln(24x) (7-x)(3x+5) = 24x -3x² + 16x + 35 = 24x -3x² - 8x + 35 = 0 (3x-7)(x+5) = 0 x = 7/3, -5 x² - 4x = log₂32 x² - 4x = 5 x² - 4x - 5 = 0 (x-5)(x+1) = 0 x = 5, -1 Write the following as a sum, difference or multiple of logarithms (EXPAND). log(√(x³/w)) = (1/2)log(x³/w) = (1/2)(log(x³) - log(w)) = (3/2)log(x) - (1/2)log(w) log(√(a³b/c⁵)) = (1/2)log(a³b/c⁵) = (1/2)(log(a³) + log(b) - log(c⁵)) = (3/2)log(a) + (1/2)log(b) - (5/2)log(c) ln(5x/∛(x²+1)) = ln(5) + ln(x) - (1/3)ln(x²+1) Write the following as the logarithm of a single quantity (CONDENSE). (1/5)[3log(x+1) + 2log(x-1) - log(7)] = (1/5)[log((x+1)³) + log((x-1)²) - log(7)] = (1/5)[log(((x+1)³(x-1)²)/7)] = log(⁵√(((x+1)³(x-1)²)/7)) (1/4)log₈16 - 2log₈5 + log₈7 = log₈16^(1/4) - log₈5² + log₈7 = log₈(2/25) + log₈7 41. If log₉9 = 3.17, find the following: a. log₉81 log₉81 = log₉9² = 2log₉9 = 2(3.17) = 6.34 b. log₉9x log₉9x = log₉9 + log₉x = 3.17 + 1 = 4.17 c. log₉9x² log₉9x² = log₉9 + log₉x² = 3.17 + 2 = 5.17 Graph the following and label the x-intercept, y-intercept, and asymptotes. Also, list the domain and range of both f(x) and g(x), as well as the interval over which it is increasing or decreasing. Lastly, list the transformations from f(x) to g(x). 1   1. 42. f(x) = 2^x; g(x) = 2^(x+1) - 3 f(x): Domain: (-∞, ∞) Range: (0, ∞) Asymptote: y = 0 Increasing: (-∞, ∞) g(x): Domain: (-∞, ∞) Range: (-3, ∞) Asymptote: y = -3 Increasing: (-∞, ∞) Transformation from f(x) to g(x): Left 1, Down 3 43. f(x) = log₃x; g(x) = log₃(-x) f(x): Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 Increasing: (0, ∞) g(x): Domain: (-∞, 0) Range: (-∞, ∞) Asymptote: x = 0 Increasing: (-∞, 0) Transformation from f(x) to g(x): Reflection over the y-axis