Exponential Logarithmic Graphs

lessonbunny.com

Answer Key & Explanations

Exponential Logarithmic Graphs · Grade 11 · Worksheet 2

1. Matiu is an environmental scientist studying the decline of a native bird population on an island. The population, P(t), after t years is modeled by the function P(t) = 1200 \times (0.85)^t, where t is time in years. Matiu needs to report the year when the population will first fall below 300 birds. Determine the number of years required for the population to reach exactly 300 birds, expressing your answer as an exact value involving logarithms. Answer: t = log(0.25) / log(0.85) Solution: Set up the equation with the given population model: 300 = 1200 × (0.85)^t Divide both sides by 1200 to isolate the exponential term: 300/1200 = (0.85)^t Simplify the fraction: 1/4 = 0.25 = (0.85)^t Take the logarithm of both sides.
   Full step-by-step solution

   Step 1: Set up the equation with the given population model: 300 = 1200 × (0.85)^t Step 2: Divide both sides by 1200 to isolate the exponential term: 300/1200 = (0.85)^t Step 3: Simplify the fraction: 1/4 = 0.25 = (0.85)^t Step 4: Take the logarithm of both sides. Using base 10 logarithms: log(0.25) = log((0.85)^t) Step 5: Use the logarithm power rule: log(0.25) = t × log(0.85) Step 6: Solve for t by dividing both sides by log(0.85): t = log(0.25) / log(0.85) Step 7: This is the exact answer expressed in terms of logarithms. The answer is t = log(0.25) / log(0.85).

2. Graph f(x) = 12^x and g(x) = log₁₂(x) on the same coordinate plane. Identify the asymptote of each function and state the domain and range of both. Answer: f(x) asymptote: y = 0; g(x) asymptote: x = 0; Domain of f: (-∞, ∞); Range of f: (0, ∞); Domain of g: (0, ∞); Range of g: (-∞, ∞) Solution: Graph f(x) = 12^x. This is an exponential function with base 12 > 1, so it increases rapidly. As x → -∞, 12^x → 0, so the horizontal asymptote is y = 0.
   Full step-by-step solution

   Step 1: Graph f(x) = 12^x. This is an exponential function with base 12 > 1, so it increases rapidly. As x → -∞, 12^x → 0, so the horizontal asymptote is y = 0. As x → ∞, 12^x → ∞. The y-intercept is (0, 1). Domain: all real numbers (-∞, ∞). Range: (0, ∞). Step 2: Graph g(x) = log₁₂(x). This is the inverse of f(x). The graph is a reflection of f(x) across the line y = x. As x → 0⁺, log₁₂(x) → -∞, so the vertical asymptote is x = 0. As x → ∞, log₁₂(x) → ∞. The x-intercept is (1, 0). Domain: (0, ∞). Range: all real numbers (-∞, ∞). Step 3: Key points for f(x): (0, 1), (1, 12), (-1, 1/12). Key points for g(x): (1, 0), (12, 1), (1/12, -1). The asymptotes are: f(x) has horizontal asymptote y = 0; g(x) has vertical asymptote x = 0.

3. log₂(64) - ln(e⁴) = ? Answer: 2 Solution: Evaluate log₂(64). Since 2^6 = 64, log₂(64) = 6. Evaluate ln(e⁴).
   Full step-by-step solution

   Step 1: Evaluate log₂(64). Since 2^6 = 64, log₂(64) = 6. Step 2: Evaluate ln(e⁴). Since ln(e⁴) = 4 × ln(e) and ln(e) = 1, ln(e⁴) = 4. Step 3: Subtract the results: 6 - 4 = 2. The answer is 2.

4. Graph f(x) = 5^x and g(x) = log_5(x) on the same coordinate plane. Identify the asymptote of each function and state the domain and range of both. Answer: f(x) asymptote: y = 0; g(x) asymptote: x = 0; Domain of f: (-∞, ∞), Range of f: (0, ∞); Domain of g: (0, ∞), Range of g: (-∞, ∞) Solution: Graph f(x) = 5^x. This is an exponential growth function. As x → -∞, 5^x → 0, so the horizontal asymptote is y = 0.
   Full step-by-step solution

   Step 1: Graph f(x) = 5^x. This is an exponential growth function. As x → -∞, 5^x → 0, so the horizontal asymptote is y = 0. As x → ∞, 5^x → ∞. The y-intercept is at (0, 1) because 5^0 = 1. Domain: all real numbers (-∞, ∞). Range: (0, ∞). Step 2: Graph g(x) = log_5(x). This is the inverse of f(x). It is defined only for x > 0, so the vertical asymptote is x = 0. As x → 0^+, log_5(x) → -∞. As x → ∞, log_5(x) → ∞. The x-intercept is at (1, 0) because log_5(1) = 0. Domain: (0, ∞). Range: all real numbers (-∞, ∞). Step 3: The graphs are reflections of each other across the line y = x, confirming they are inverses. Final answer: f(x) asymptote: y = 0; g(x) asymptote: x = 0; Domain of f: (-∞, ∞), Range of f: (0, ∞); Domain of g: (0, ∞), Range of g: (-∞, ∞).

5. Sophia is a marine biologist studying the light penetration in a coastal ocean. The intensity of light, I(x), in lumens per square meter, at a depth of x meters below the surface is modeled by the function I(x) = 16 × (1/2)^(x/6). At what depth, in meters, will the light intensity be 1 lumen per square meter? Express your answer as an exact value using logarithms, then evaluate to the nearest tenth of a meter. Answer: 24 meters Solution: Set up the equation with the given intensity: 1 = 16 × (1/2)^(x/6) Divide both sides by 16: 1/16 = (1/2)^(x/6) Recognize that 1/16 = (1/2)^4 because 2^4 = 16, so (1/2)^4 = 1/16 So (1/2)^4 = (1/2)^(x/6) Since the bases are equal (both 1/2), set the exponents equal: 4 = x/6 Multiply both sides by…
   Full step-by-step solution

   Step 1: Set up the equation with the given intensity: 1 = 16 × (1/2)^(x/6) Step 2: Divide both sides by 16: 1/16 = (1/2)^(x/6) Step 3: Recognize that 1/16 = (1/2)^4 because 2^4 = 16, so (1/2)^4 = 1/16 Step 4: So (1/2)^4 = (1/2)^(x/6) Step 5: Since the bases are equal (both 1/2), set the exponents equal: 4 = x/6 Step 6: Multiply both sides by 6: x = 24 Step 7: Alternatively, using logarithms: take log base 1/2 of both sides: log_{1/2}(1/16) = x/6, which gives 4 = x/6, so x = 24. The answer is 24 meters.

6. Graph f(x) = 8^x and g(x) = log_8(x) on the same coordinate plane. Identify the asymptote of each function and state the domain and range of both. Answer: f(x) = 8^x: asymptote y = 0, domain (-∞, ∞), range (0, ∞); g(x) = log_8(x): asymptote x = 0, domain (0, ∞), range (-∞, ∞) Solution: Graph f(x) = 8^x. Since the base 8 > 1, the function is increasing. As x → -∞, 8^x → 0, so the horizontal asymptote is y = 0.
   Full step-by-step solution

   Step 1: Graph f(x) = 8^x. Since the base 8 > 1, the function is increasing. As x → -∞, 8^x → 0, so the horizontal asymptote is y = 0. The y-intercept is at (0, 1) because 8^0 = 1. Domain: all real numbers (-∞, ∞). Range: (0, ∞). Step 2: Graph g(x) = log_8(x). This is the inverse of f(x) = 8^x. The graph is reflected across the line y = x. As x → 0^+, log_8(x) → -∞, so the vertical asymptote is x = 0. The x-intercept is at (1, 0) because log_8(1) = 0. Domain: (0, ∞). Range: all real numbers (-∞, ∞). Step 3: Key points for f(x): (-1, 1/8), (0, 1), (1, 8). Key points for g(x): (1/8, -1), (1, 0), (8, 1). The graphs are symmetric about y = x. The answer is: f(x) = 8^x has asymptote y = 0, domain (-∞, ∞), range (0, ∞); g(x) = log_8(x) has asymptote x = 0, domain (0, ∞), range (-∞, ∞).

7. Noah is a seismologist analyzing the decay of seismic wave energy as it travels through the Earth's crust. The energy E(t) in joules of a particular seismic wave at time t seconds is modeled by the function E(t) = 16 × 2^(-t/6). Graph the function E(t) over the domain t ≥ 0, and identify the horizontal asymptote and the y-intercept. Then, determine after how many seconds the energy will first drop below 1 joule. Express your answer as an exact value using logarithms. Answer: 24 seconds Solution: Identify key features of the graph. The function is E(t) = 16 × 2^(-t/6). This is an exponential decay function.
   Full step-by-step solution

   Step 1: Identify key features of the graph. The function is E(t) = 16 × 2^(-t/6). This is an exponential decay function. The y-intercept occurs at t = 0: E(0) = 16 × 2^0 = 16. So the y-intercept is (0, 16). The horizontal asymptote is the line y = 0, because as t → ∞, 2^(-t/6) → 0, so E(t) → 0. Step 2: To find when E(t) < 1, set up the inequality: 16 × 2^(-t/6) < 1. Divide both sides by 16: 2^(-t/6) < 1/16. Note that 1/16 = 2^(-4). So the inequality becomes 2^(-t/6) < 2^(-4). Step 3: Since the base 2 is greater than 1, the inequality direction is preserved when comparing exponents: -t/6 < -4. Multiply both sides by -6 (reversing the inequality): t > 24. Step 4: Therefore, the energy drops below 1 joule after t = 24 seconds. The exact answer is 24 seconds. The answer is 24 seconds.