Exponential Logarithmic Graphs

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Answer Key & Explanations

Exponential Logarithmic Graphs · Grade 11 · Worksheet 1

1. Olivia is investigating the spread of a rare plant species in a protected wetland. The area covered by the plant, in square meters, is modeled by the function A(t) = 7 × 3^(t/5), where t is the number of years since the study began. Emma, her colleague, is studying a different plant species whose area is modeled by B(t) = 21 × 3^(t/5). Emma claims that her graph is a vertical shift of Olivia's graph. Is Emma correct? If not, describe the correct transformation from A(t) to B(t), identify the horizontal asymptote of each function, and state the y-intercept of A(t). Answer: Emma is not correct. The transformation is a vertical stretch by a factor of 3. The horizontal asymptote of both functions is y = 0. The y-intercept of A(t) is 7. Solution: Compare the two functions: A(t) = 7 × 3^(t/5) and B(t) = 21 × 3^(t/5). Notice that 21 = 3 × 7, so B(t) = 3 × (7 × 3^(t/5)) = 3 × A(t). This is a vertical stretch by a factor of 3, not a vertical shift.
   Full step-by-step solution

   Step 1: Compare the two functions: A(t) = 7 × 3^(t/5) and B(t) = 21 × 3^(t/5). Step 2: Notice that 21 = 3 × 7, so B(t) = 3 × (7 × 3^(t/5)) = 3 × A(t). This is a vertical stretch by a factor of 3, not a vertical shift. Emma is incorrect. Step 3: For both functions, as t → -∞, 3^(t/5) → 0, so both A(t) and B(t) approach 0. The horizontal asymptote for both is y = 0. Step 4: The y-intercept occurs when t = 0. For A(0) = 7 × 3^(0/5) = 7 × 3^0 = 7 × 1 = 7. So the y-intercept is 7. The answer is: Emma is not correct. The transformation is a vertical stretch by a factor of 3. The horizontal asymptote of both functions is y = 0. The y-intercept of A(t) is 7.

2. Graph f(x) = 3^(x+1) - 2 and identify the horizontal asymptote. Answer: y = -2 Solution: Start with the base exponential function f(x) = 3^x. Its horizontal asymptote is y = 0. The function is f(x) = 3^(x+1) - 2.
   Full step-by-step solution

   Step 1: Start with the base exponential function f(x) = 3^x. Its horizontal asymptote is y = 0. Step 2: The function is f(x) = 3^(x+1) - 2. The (x+1) inside the exponent shifts the graph 1 unit to the left, but does not affect the asymptote. Step 3: The -2 outside the exponent shifts the entire graph downward by 2 units. Step 4: Therefore, the horizontal asymptote shifts from y = 0 to y = -2. Step 5: To verify, as x approaches negative infinity, 3^(x+1) approaches 0, so f(x) approaches -2. The horizontal asymptote is y = -2.

3. A radioactive substance decays according to the function N(t) = 100e^(-0.0231t), where N is the amount remaining in grams and t is time in years. The graph of this function shows exponential decay. After how many years will exactly half of the original 100 grams remain? Answer: 30 Solution: N(t) = 100 * e^(-0.0231 * t) We want the time t when exactly half of the original 100 grams remains. Half of 100 grams is 50 grams.
   Full step-by-step solution

   We are given the decay function: N(t) = 100 * e^(-0.0231 * t) We want the time t when exactly half of the original 100 grams remains. Half of 100 grams is 50 grams. So we set N(t) = 50: 50 = 100 * e^(-0.0231 * t) Step 1: Divide both sides by 100: 50/100 = e^(-0.0231 * t) 0.5 = e^(-0.0231 * t) Step 2: Take the natural logarithm of both sides: ln(0.5) = ln(e^(-0.0231 * t)) Step 3: Use the property ln(e^x) = x: ln(0.5) = -0.0231 * t Step 4: Solve for t: t = ln(0.5) / (-0.0231) Step 5: Calculate ln(0.5): ln(0.5) ≈ -0.693147 So: t = (-0.693147) / (-0.0231) t ≈ 0.693147 / 0.0231 Step 6: Perform the division: 0.693147 / 0.0231 ≈ 30.002 Step 7: Round to the nearest whole number (since the problem likely expects an integer number of years): t ≈ 30 years Thus, after 30 years, exactly half of the original substance remains.

4. Graph f(x) = 3^(x+1) - 2. Identify the horizontal asymptote and the y-intercept. Answer: Horizontal asymptote: y = -2; y-intercept: (0, 1) Solution: Start with the parent function f(x) = 3^x. Its horizontal asymptote is y = 0. The transformation f(x) = 3^(x+1) - 2 involves a horizontal shift left by 1 unit and a vertical shift down by 2 units.
   Full step-by-step solution

   Step 1: Start with the parent function f(x) = 3^x. Its horizontal asymptote is y = 0. Step 2: The transformation f(x) = 3^(x+1) - 2 involves a horizontal shift left by 1 unit and a vertical shift down by 2 units. Step 3: The horizontal asymptote shifts with the vertical translation: from y = 0 to y = -2. Step 4: To find the y-intercept, set x = 0: f(0) = 3^(0+1) - 2 = 3^1 - 2 = 3 - 2 = 1. So the y-intercept is (0, 1). Step 5: The graph passes through (0, 1), approaches y = -2 as x → -∞, and increases without bound as x → +∞. The answer is: Horizontal asymptote: y = -2; y-intercept: (0, 1).

5. log₃(81) - 2² = ? Answer: 0 Solution: Evaluate log₃(81). Since 3⁴ = 81, log₃(81) = 4. Evaluate 2² = 4.
   Full step-by-step solution

   Step 1: Evaluate log₃(81). Since 3⁴ = 81, log₃(81) = 4. Step 2: Evaluate 2² = 4. Step 3: Subtract the results: 4 - 4 = 0. The answer is 0.

6. Noah is a financial analyst tracking the growth of a technology stock. The value of the stock, V(t), in dollars, is modeled by the function V(t) = 120 × 3^(t/7), where t is the number of months since the stock was first listed. He needs to create a graph of this function for his report. Identify the horizontal asymptote of this function, state the y-intercept, and determine the time, in months, when the stock value will reach $3,240. Round your answer to the nearest tenth of a month. Answer: t = 14.0 months Solution: Identify the horizontal asymptote. For V(t) = 120 × 3^(t/7), as t approaches negative infinity, 3^(t/7) approaches 0, so V(t) approaches 0. The horizontal asymptote is y = 0.
   Full step-by-step solution

   Step 1: Identify the horizontal asymptote. For V(t) = 120 × 3^(t/7), as t approaches negative infinity, 3^(t/7) approaches 0, so V(t) approaches 0. The horizontal asymptote is y = 0. Step 2: Find the y-intercept. Evaluate V(0) = 120 × 3^(0/7) = 120 × 3^0 = 120 × 1 = 120. The y-intercept is (0, 120). Step 3: Set up the equation for V(t) = 3240. 3240 = 120 × 3^(t/7) Step 4: Divide both sides by 120: 3240/120 = 3^(t/7). Simplify: 27 = 3^(t/7) Step 5: Write 27 as a power of 3: 27 = 3^3. So, 3^3 = 3^(t/7) Step 6: Since the bases are equal, set exponents equal: 3 = t/7 Step 7: Multiply both sides by 7: t = 21. This is exact, but the problem asks to round to the nearest tenth: t = 21.0 months. The answer is t = 21.0 months. (Note: The problem originally set V(t)=3240, but 3240/120 = 27, so t=21 exactly. The rounding instruction is given for consistency with problem format.)

7. Graph f(x) = 5^x and g(x) = log_5(x) on the same coordinate plane. Identify the asymptote of f(x) and the domain of g(x). Answer: Asymptote of f(x): y = 0; Domain of g(x): x > 0 Solution: Graph f(x) = 5^x. This is an exponential growth function. Create a table of values: x = -2 gives 5^(-2) = 1/25 = 0.04; x = -1 gives 5^(-1) = 0.2; x = 0 gives 5^0 = 1; x = 1 gives 5^1 = 5; x = 2 gives 5^2 = 25.
   Full step-by-step solution

   Step 1: Graph f(x) = 5^x. This is an exponential growth function. Create a table of values: x = -2 gives 5^(-2) = 1/25 = 0.04; x = -1 gives 5^(-1) = 0.2; x = 0 gives 5^0 = 1; x = 1 gives 5^1 = 5; x = 2 gives 5^2 = 25. Plot these points and draw a smooth curve. The horizontal asymptote is y = 0 because as x approaches negative infinity, 5^x approaches 0 but never reaches it. Step 2: Graph g(x) = log_5(x). Since log_5(x) is the inverse of 5^x, the graph is the reflection of f(x) across the line y = x. Create a table of values: x = 1/25 gives log_5(1/25) = -2; x = 1/5 gives log_5(1/5) = -1; x = 1 gives log_5(1) = 0; x = 5 gives log_5(5) = 1; x = 25 gives log_5(25) = 2. Plot these points and draw a smooth curve. The vertical asymptote is x = 0. Step 3: Identify key features. For f(x) = 5^x, the horizontal asymptote is y = 0. For g(x) = log_5(x), the domain is all positive real numbers, written as x > 0 or (0, infinity). The answer is: Asymptote of f(x): y = 0; Domain of g(x): x > 0.