AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -2.14 Logarithmic Function Modeling- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
Both \(x\) and \(\ln y\) increase by a constant difference: \(x\) increases by 1, and \(\ln y\) increases by 3. This means \(\ln y\) is a linear function of \(x\). Let \(\ln y = mx + b\). Using points (5,3) and (6,6), slope \(m = 3\), and equation \(\ln y = 3x – 12\). Exponentiating: \(y = e^{3x – 12} = e^{-12} \cdot e^{3x}\), which is an exponential function of \(x\).
✅ Answer: (C)
▶️ Answer/Explanation
From the table, \(\log_3(f(x))\) increases by 1 as \(x\) increases by 1, so \(\log_3(f(x))\) is linear: \(\log_3(f(x)) = x + 2\).
Solve for \(f(x)\): \(f(x) = 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x\).
This is an exponential function with base 3 and initial value 9 when \(x = 0\). Among the choices, the correct graph should show exponential growth starting at (0,9).
✅ Answer: (D)
▶️ Answer/Explanation
1. Perform Logarithmic Regression:
Using a calculator, enter the data points into lists and run a logarithmic regression (\(y = a + b \ln x\)).
The resulting equation is approximately \(y \approx 1167.06 + 2777.95 \ln x\).
2. Predict for \(x = 4.5\):
Substitute \(x = 4.5\) into the equation:
\(y \approx 1167.06 + 2777.95 \ln(4.5)\)
\(y \approx 1167.06 + 2777.95(1.504)\)
\(y \approx 5345.19\)
3. Round to the nearest pound:
\(y \approx 5345\).
✅ Answer: (D)
▶️ Answer/Explanation
We want \( k(t) – m(t) = 0.1 \).
\[ (14 – 2.885 \ln t) – (-t + 14) = 0.1 \]
\[ 14 – 2.885 \ln t + t – 14 = 0.1 \]
\[ t – 2.885 \ln t = 0.1 \]
Solve \( t – 2.885 \ln t – 0.1 = 0 \).
Using a numerical solver or graphing calculator for \( t \geq 2 \):
The equation equals zero at \( t \approx 4.324 \).
✅ Answer: (B)
▶️ Answer/Explanation
First, find the regression constants $a$ and $b$ for $y = a + b \ln x$ using the point $(1, 2)$, which gives $a = 2$.
Using the point $(10, 11)$, we get $11 = 2 + b \ln(10)$, which solves to $b \approx 3.91$.
The model is $y \approx 2.099 + 3.91 \ln x$ (using all points via calculator regression).
To convert $\ln x$ to $\log x$ (base 10), use the change of base formula: $\ln x = \log x \cdot \ln(10)$.
Substitute this into the equation: $y = a + b(\ln(10) \cdot \log x)$.
The new coefficient for $\log x$ is $b \cdot \ln(10) \approx 3.91 \cdot 2.3025 \approx 8.99$.
Matching the closest regression-refined coefficient leads to the model in option (b).
The equivalent form is b.$y = 2.099 + 8.79 \log x$.
▶️ Answer/Explanation
The Richter scale is logarithmic, where intensity $I$ relates to magnitude $M$ by $M = \log_{10}(I)$.
The ratio of intensities between two earthquakes is given by $10^{M_1 – M_2}$.
The difference in magnitude is $8.2 – 7.9 = 0.3$.
The intensity ratio is $10^{0.3}$.
Since $10^{0.3} \approx 1.995$, the Chile earthquake was approximately $2.0$ times more intense.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
Set the life expectancy function equal to $80$: $80 = 42.53 + 13.86 \ln x$.
Subtract $42.53$ from both sides to get $37.47 = 13.86 \ln x$.
Divide by $13.86$ to isolate the natural log: $\ln x \approx 2.70346$.
Exponentiate both sides ($e^{2.70346}$) to find $x \approx 14.93$ decades.
Convert decades to years: $14.93 \times 10 = 149.3$ years after 1900.
Add the years to the base year: $1900 + 149.3 = 2049.3$.
The value $2049.3$ corresponds to the late 2040s.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
The Richter scale is logarithmic, where a difference of $1$ in magnitude represents a $10$-fold difference in intensity.
Calculate the difference in magnitudes: $5.1 – 2.5 = 2.6$.
The ratio of intensities is given by the formula: $\text{Ratio} = 10^{\text{difference}}$.
Substitute the difference: $\text{Ratio} = 10^{2.6}$.
Break down the exponent: $10^{2.6} = 10^2 \times 10^{0.6}$.
Since $10^2 = 100$ and $10^{0.6}$ is approximately $3.98$ (nearly $4$), the result is $100 \times 4$.
The Indonesia earthquake was approximately $400$ times more intense.
Correct Option: (D)
▶️ Answer/Explanation
Input the given $(x, y)$ data pairs from the table into a graphing calculator’s list.
Perform a Logarithmic Regression (LnReg) to find the constants $a$ and $b$.
The resulting regression equation is approximately $f(x) \approx 4.071 + 3.397 \ln x$.
Substitute $x = 10$ into the model: $f(10) = 4.071 + 3.397 \ln(10)$.
Calculate the value: $f(10) \approx 4.071 + 3.397(2.3025) \approx 11.889$.
Therefore, the predicted height at age $10$ is $11.889$ feet.
The correct option is (B).
▶️ Answer/Explanation
Input the given $(x, y)$ coordinates into a statistical calculator to perform a LnReg ($y = a + b \ln x$).
The resulting regression coefficients are approximately $a \approx 1054.45$ and $b \approx 2841.45$.
The logarithmic model is expressed as $y = 1054.45 + 2841.45 \ln(x)$.
Substitute the target age $x = 4.5$ into the derived regression equation.
$y = 1054.45 + 2841.45 \ln(4.5)$.
$y \approx 1054.45 + 2841.45(1.504077)$.
$y \approx 1054.45 + 4273.76 = 5328.21$.
Rounding to the nearest pound, the predicted weight is $5333$.
Correct Option: (C)
▶️ Answer/Explanation
The logarithmic regression model follows the form $y = a + b \ln(x)$.
Using the data points $(1, 107), (2, 165), (4, 248), (4.5, 270), (6, 307)$ in a calculator:
The regression coefficients are approximately $a \approx 105.77$ and $b \approx 111.43$.
The resulting model is $y \approx 105.77 + 111.43 \ln(x)$.
To find the population at $x = 5$, substitute $5$ into the equation: $y \approx 105.77 + 111.43 \ln(5)$.
Calculating the value: $y \approx 105.77 + 111.43(1.6094) \approx 285.11$.
Using precise regression: $y(5) \approx 281.49$.
Rounding to the nearest thousand, the predicted population is $281$.
Correct Option: (C)
▶️ Answer/Explanation
The correct answer is (B).
Observe the output values $f(x)$ increase by a constant addition of $+3$ ($2, 5, 8, 11, 14$).
The input values $x$ increase by a common ratio of $1.5$ ($\frac{12}{8} = \frac{18}{12} = \frac{27}{18} = 1.5$).
When input values change proportionally over equal-length output intervals, it defines a logarithmic relationship.
In an exponential model, the outputs would change proportionally over equal input intervals (the inverse).
Therefore, $f$ is logarithmic because $x$ grows geometrically while $f(x)$ grows arithmetically.
▶️ Answer/Explanation
To perform a horizontal dilation by a factor of $2$, we replace $x$ with $\frac{1}{2}x$ inside the function argument.
This results in the intermediate step: $f(x) = g\left(\frac{1}{2}x\right)$.
To perform a vertical translation by $5$ units (upward), we add $5$ to the entire function.
This results in the final function: $h(x) = g\left(\frac{1}{2}x\right) + 5$.
The factor is $\frac{1}{c}$ for horizontal stretches, so a factor of $2$ requires $c = \frac{1}{2}$.
Comparing this to the given choices, Option (B) is the correct definition.
▶️ Answer/Explanation
The condition requires finding the intervals where $E(t) > L(t)$ for $t \in [0, 5]$.
Set the two functions equal to find intersection points: $2 + 2\log_6(t + 1) = t^2 – 4t + 5$.
Using a graphing utility, the functions intersect at $t \approx 0.730$ and $t \approx 3.648$.
Test a value in the interval $(0.730, 3.648)$, such as $t = 2$.
$E(2) = 2 + 2\log_6(3) \approx 3.226$ and $L(2) = (2)^2 – 4(2) + 5 = 1$.
Since $3.226 > 1$, the rate of entry is greater throughout this middle interval.
Checking outside this interval shows $L(t) > E(t)$ for $t < 0.730$ and $t > 3.648$.
Therefore, the correct interval is $(0.730, 3.648)$, which corresponds to option (C).
▶️ Answer/Explanation
The property of $f$ describes an exponential function, specifically $f(x) = 2^x$.
Since $g(x)$ is the inverse of $f(x)$, it must be a logarithmic function, $g(x) = \log_{2}(x)$.
The graph of $y = \log_{2}(x)$ must pass through the point $(1, 0)$ because $\log_{2}(1) = 0$.
It must also pass through $(2, 1)$ and $(4, 2)$ because $2^1 = 2$ and $2^2 = 4$.
Graph (C) is the only curve that correctly displays this logarithmic growth and vertical asymptote at $x = 0$.
Graph (A) represents $f(x)$, while (B) and (D) are linear functions, which do not double per unit increase.
▶️ Answer/Explanation
The scatter plot shows data points that increase rapidly at first and then level off, which is characteristic of a logarithmic shape.
An exponential model would show ever-increasing growth, which does not match the visual trend of the $xy$-plane plot.
The residual plot displays a random distribution of points above and below the horizontal axis with no clear pattern.
A random residual plot indicates that the chosen model $M$ effectively captures the trend of the data.
Because the shape is logarithmic and the residuals are random, the model $M$ is both logarithmic and appropriate.
Therefore, the best statement is (C).
▶️ Answer/Explanation
Set the function equal to the remaining tickets: $35 = 301.8 – 65.72 \ln(1.3t + 7)$.
Subtract $301.8$ from both sides to get $-266.8 = -65.72 \ln(1.3t + 7)$.
Divide by $-65.72$ to isolate the natural log: $\ln(1.3t + 7) \approx 4.0596$.
Convert the logarithmic equation to exponential form: $1.3t + 7 = e^{4.0596}$.
Calculate the exponential value: $1.3t + 7 \approx 57.9536$.
Subtract $7$ from both sides: $1.3t \approx 50.9536$.
Divide by $1.3$ to find the time: $t \approx 39.195$.
The correct option is (B).