Exponential Logarithmic Graphs

lessonbunny.com

Answer Key & Explanations

Exponential Logarithmic Graphs · Grade 11 · Worksheet 1

1. log₂(64) + ln(e⁵) = ? Answer: 11 Solution: Evaluate log₂(64). Since 2^6 = 64, log₂(64) = 6. Evaluate ln(e⁵).
   Full step-by-step solution

   Step 1: Evaluate log₂(64). Since 2^6 = 64, log₂(64) = 6. Step 2: Evaluate ln(e⁵). Since ln is log base e, and e^5 = e^5, ln(e⁵) = 5. Step 3: Add the results: 6 + 5 = 11. The answer is 11.

2. On a coordinate plane, the graph of the function f(x) = 2^x is shown. This exponential curve passes through point A at (3, 8). The graph is then reflected across the line y = x to create the inverse function g(x). What is the y-coordinate of the point on g(x) that corresponds to x = 8? Answer: 3 Solution: The original function is f(x) = 2^x, which passes through (3, 8). When a function is reflected across the line y = x, we get its inverse function.
   Full step-by-step solution

   Step 1: The original function is f(x) = 2^x, which passes through (3, 8). Step 2: When a function is reflected across the line y = x, we get its inverse function. Step 3: For inverse functions, if (a, b) is on f(x), then (b, a) is on the inverse function g(x). Step 4: Since (3, 8) is on f(x), then (8, 3) must be on g(x). Step 5: Therefore, when x = 8 on g(x), the y-coordinate is 3. The answer is 3.

3. Graph f(x) = 5^(x+2) - 3. Identify the horizontal asymptote and the y-intercept. Answer: Horizontal asymptote: y = -3; y-intercept: (0, 22) Solution: Start with the parent function f(x) = 5^x. Its horizontal asymptote is y = 0 and y-intercept is (0, 1). The +2 inside the exponent shifts the graph left by 2 units.
   Full step-by-step solution

   Step 1: Start with the parent function f(x) = 5^x. Its horizontal asymptote is y = 0 and y-intercept is (0, 1). Step 2: Apply the transformation f(x) = 5^(x+2) - 3. The +2 inside the exponent shifts the graph left by 2 units. The -3 outside shifts the graph down by 3 units. Step 3: The horizontal asymptote of 5^x is y = 0. Shifting down by 3 gives the new asymptote: y = -3. Step 4: Find the y-intercept by setting x = 0: f(0) = 5^(0+2) - 3 = 5^2 - 3 = 25 - 3 = 22. So the y-intercept is (0, 22). Step 5: The graph is an increasing exponential curve approaching y = -3 from above as x goes to negative infinity, passing through (0, 22). Final answer: Horizontal asymptote: y = -3; y-intercept: (0, 22).

4. log₃(81) - 2log₃(9) + log₃(1) = ? Answer: 0 Solution: Evaluate log₃(81). Since 3⁴ = 81, log₃(81) = 4. Evaluate 2log₃(9).
   Full step-by-step solution

   Step 1: Evaluate log₃(81). Since 3⁴ = 81, log₃(81) = 4. Step 2: Evaluate 2log₃(9). Since 3² = 9, log₃(9) = 2, so 2 × 2 = 4. Step 3: Evaluate log₃(1). Since 3⁰ = 1, log₃(1) = 0. Step 4: Substitute back into the expression: 4 - 4 + 0 = 0. The answer is 0.

5. Ava is a civil engineer monitoring the structural integrity of a bridge. The load-bearing capacity C(t) in megapascals of a support beam over time t in years is modeled by the function C(t) = 26 × 0.96^(t/2). Ava needs to determine when the beam's capacity will drop to 13 megapascals, which is the safety threshold for replacement. Write an equation representing this situation and solve for the time t in years, expressing your answer in terms of logarithms (use base-10 or natural log). Answer: t = 2 × log(0.5)/log(0.96) years (or t = 2 × ln(0.5)/ln(0.96) years) Solution: Set up the equation: 13 = 26 × 0.96^(t/2) Divide both sides by 26: 13/26 = 0.96^(t/2) Simplify the fraction: 1/2 = 0.96^(t/2) Take the logarithm of both sides (using base-10): log(1/2) = log(0.96^(t/2)) Use the power rule of logarithms: log(1/2) = (t/2) × log(0.96) Multiply both sides by 2: 2 ×…
   Full step-by-step solution

   Step 1: Set up the equation: 13 = 26 × 0.96^(t/2) Step 2: Divide both sides by 26: 13/26 = 0.96^(t/2) Step 3: Simplify the fraction: 1/2 = 0.96^(t/2) Step 4: Take the logarithm of both sides (using base-10): log(1/2) = log(0.96^(t/2)) Step 5: Use the power rule of logarithms: log(1/2) = (t/2) × log(0.96) Step 6: Multiply both sides by 2: 2 × log(1/2) = t × log(0.96) Step 7: Divide both sides by log(0.96): t = 2 × log(1/2) / log(0.96) Step 8: Note that log(1/2) = log(2^(-1)) = -log(2), so t = 2 × (-log(2)) / log(0.96) = -2 × log(2)/log(0.96) Step 9: Alternatively, using natural logs: t = 2 × ln(1/2)/ln(0.96) = 2 × (-ln(2))/ln(0.96) = -2 × ln(2)/ln(0.96) The answer is t = 2 × log(1/2)/log(0.96) years (or equivalently t = -2 × log(2)/log(0.96) years).

6. A radioactive substance decays according to the function N(t) = 500e^(-0.0231t), where N is the number of atoms remaining and t is time in years. On a coordinate plane, the exponential decay curve passes through the point (30, N(30)). What is the half-life of this substance, rounded to the nearest year? Answer: 30 Solution: N(t) = 500 e^(-0.0231 t) Half-life is the time t such that N(t) = N(0)/2. Initial amount N(0) = 500 e^(0) = 500. Half of that is 250.
   Full step-by-step solution

   Step 1: Understand the problem We are given the decay function: N(t) = 500 e^(-0.0231 t) Half-life is the time t such that N(t) = N(0)/2. Step 2: Set up the half-life equation Initial amount N(0) = 500 e^(0) = 500. Half of that is 250. So we solve: 250 = 500 e^(-0.0231 t) Step 3: Solve for t Divide both sides by 500: 250/500 = e^(-0.0231 t) 1/2 = e^(-0.0231 t) Step 4: Take natural logarithm of both sides ln(1/2) = -0.0231 t ln(1) - ln(2) = -0.0231 t 0 - ln(2) = -0.0231 t - ln(2) = -0.0231 t Step 5: Cancel the negative signs ln(2) = 0.0231 t Step 6: Solve for t t = ln(2) / 0.0231 Step 7: Compute numerical value ln(2) ≈ 0.693147 t ≈ 0.693147 / 0.0231 ≈ 30.006 Step 8: Round to nearest year t ≈ 30 years Thus, the half-life is 30 years.

7. Graph f(x) = 2^(x-3) - 2. Identify the horizontal asymptote and the y-intercept. Answer: Horizontal asymptote: y = -2; y-intercept: (0, -15/8) Solution: Start with the parent function f(x) = 2^x. Its horizontal asymptote is y = 0. The function is f(x) = 2^(x-3) - 2.
   Full step-by-step solution

   Step 1: Start with the parent function f(x) = 2^x. Its horizontal asymptote is y = 0. Step 2: The function is f(x) = 2^(x-3) - 2. The term (x-3) shifts the graph 3 units to the right. The -2 shifts the graph 2 units downward. Step 3: The horizontal asymptote of the parent function is y = 0. After a vertical shift of -2, the new horizontal asymptote is y = -2. Step 4: To find the y-intercept, set x = 0: f(0) = 2^(0-3) - 2 = 2^(-3) - 2 = 1/8 - 2 = 1/8 - 16/8 = -15/8. Step 5: The y-intercept is at (0, -15/8). The answer is: Horizontal asymptote: y = -2; y-intercept: (0, -15/8).

8. 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.