▶️ Answer/Explanation
The domain of a natural logarithm $\ln(u)$ requires the argument to be strictly positive: $2x^2 – 3x – 20 > 0$.
Factor the quadratic expression: $(2x + 5)(x – 4) > 0$.
Identify the critical values (roots) where the expression equals zero: $x = -\frac{5}{2}$ and $x = 4$.
Test intervals on a number line: $(-\infty, -\frac{5}{2})$, $(-\frac{5}{2}, 4)$, and $(4, \infty)$.
The expression is positive when $x < -\frac{5}{2}$ or $x > 4$.
In interval notation, the domain is $(-\infty, -\frac{5}{2}) \cup (4, \infty)$.
Therefore, the correct option is A.
▶️ Answer/Explanation
Use the product property of logarithms: $\log_{4}[x(x – 6)] = 2$.
Convert to exponential form: $x(x – 6) = 4^2$, which simplifies to $x^2 – 6x = 16$.
Rearrange into a quadratic equation: $x^2 – 6x – 16 = 0$.
Factor the quadratic: $(x – 8)(x + 2) = 0$, giving potential solutions $x = 8$ and $x = -2$.
Check for extraneous solutions: $x$ must be greater than $6$ for the original logarithms to be defined.
Since $\log_{4}(-2)$ is undefined, $x = -2$ is extraneous.
The only valid solution is $x = 8$.
Correct Option: D
▶️ Answer/Explanation
From the graph, identify the first two terms: $a_1 = 12$ and $a_2 = 6$.
Calculate the common ratio $r$ using $r = \frac{a_2}{a_1} = \frac{6}{12} = \frac{1}{2}$.
The standard formula for a geometric sequence is $a_n = a_1(r)^{n-1}$.
Substituting the values gives $a_n = 12\left(\frac{1}{2}\right)^{n-1}$.
To match Option B, rewrite the expression: $12\left(\frac{1}{2}\right)^{n-1} = 6 \cdot 2 \cdot \left(\frac{1}{2}\right)^{n-1}$.
Since $2 = \left(\frac{1}{2}\right)^{-1}$, the expression becomes $6\left(\frac{1}{2}\right)^{-1}\left(\frac{1}{2}\right)^{n-1}$.
Simplify the exponents to get $a_n = 6\left(\frac{1}{2}\right)^{n-2}$.
Therefore, the correct choice is B.
▶️ Answer/Explanation
The correct answer is C.
Start with the given relationship between hours and days: $t = 24d$.
Solve for $d$ to express days in terms of hours: $d = \frac{t}{24}$.
Substitute this expression for $d$ into the original function: $P(d) = A_0(0.5)^{\frac{d}{8}}$.
This gives $H(t) = A_0(0.5)^{\frac{t/24}{8}}$.
Simplify the exponent by multiplying the denominators: $24 \times 8 = 192$.
The resulting function is $H(t) = A_0(0.5)^{t/192}$.
▶️ Answer/Explanation
The correct option is B.
Rewrite the radical as an exponent: $\ln (x – 5)^{1/2} = 4$.
Apply the power rule for logarithms: $\frac{1}{2} \ln (x – 5) = 4$.
Multiply both sides by 2 to isolate the log: $\ln (x – 5) = 8$.
Convert the natural log to exponential form (base $e$): $x – 5 = e^{8}$.
Solve for $x$ by adding 5 to both sides: $x = e^{8} + 5$.
▶️ Answer/Explanation
The correct answer is B.
Let $u = 2^x$, then $f(x)$ becomes $\frac{u^2 – 14u + 49}{u – 7} = \frac{(u-7)^2}{u-7}$.
For $u \neq 7$, the function simplifies to $f(x) = 2^x – 7$.
As $x \to -\infty$, $2^x \to 0$, so $f(x) \to 0 – 7 = -7$, making statement I true.
At $x = \log_2 7$, $u = 7$, creating a $\frac{0}{0}$ indeterminate form.
Since the factor $(2^x – 7)$ cancels out, $x = \log_2 7$ is a hole, not an asymptote.
Therefore, statement II is true and statement III is false.
▶️ Answer/Explanation
Step 1: Identify the domain constraint: the argument of the log must be positive, so $x + 3 > 0$, which means $x > -3$.
Step 2: Rewrite the logarithmic inequality in exponential form: $x + 3 \leq 2^{4}$.
Step 3: Simplify the power: $x + 3 \leq 16$.
Step 4: Solve for $x$: subtracting $3$ from both sides gives $x \leq 13$.
Step 5: Combine the domain constraint ($x > -3$) with the inequality solution ($x \leq 13$).
Step 6: The final solution set is $-3 < x \leq 13$.
Correct Option: D
▶️ Answer/Explanation
The property “input doubles, output decreases by $1$” describes a logarithmic relationship.
Mathematically, this can be expressed as $h(2x) = h(x) – 1$.
Checking coordinates on Graph C: at $x = 1$, $y = 4$.
When the input doubles to $x = 2$, the output is $y = 3$ (a decrease of $1$).
When the input doubles again to $x = 4$, the output is $y = 2$ (a decrease of $1$).
When the input doubles to $x = 8$, the output is $y = 1$ (a decrease of $1$).
Therefore, Graph C is the only one that satisfies this specific logarithmic decay.
▶️ Answer/Explanation
The formula for the \(n^{th}\) term of a geometric sequence is \(a_n = a_1 \cdot r^{n-1}\).
Substitute the given values: \(a_1 = -\frac{1}{2}\), \(r = \frac{1}{3}\), and \(n = 8\).
Calculate the exponent: \(n – 1 = 8 – 1 = 7\).
Evaluate the power: \((\frac{1}{3})^7 = \frac{1}{2187}\).
Multiply by the first term: \(a_8 = -\frac{1}{2} \cdot \frac{1}{2187}\).
Final result: \(a_8 = -\frac{1}{4374}\).
Therefore, the correct option is A.
▶️ Answer/Explanation
The first term of the arithmetic sequence is $a_1 = 14.2$.
The common difference is $d = 23.8 – 14.2 = 9.6$.
The formula for the $n^{th}$ term is $a_n = a_1 + (n – 1)d$.
Substitute $n = 14$ into the formula: $a_{14} = 14.2 + (14 – 1)(9.6)$.
Simplify the expression: $a_{14} = 14.2 + (13)(9.6)$.
Calculate the product: $13 \times 9.6 = 124.8$.
Find the final sum: $14.2 + 124.8 = 139.0$.
Therefore, the object will fall $139.0$ meters in $14$ seconds.
▶️ Answer/Explanation
Set the demand equation to the target value: $225,000 = 54,000 \ln(5t + 3)$.
Divide both sides by $54,000$: $\frac{225,000}{54,000} = \ln(5t + 3)$, which simplifies to $4.1667 \approx \ln(5t + 3)$.
Convert from logarithmic to exponential form: $e^{4.1667} = 5t + 3$.
Calculate the exponential value: $64.501 \approx 5t + 3$.
Subtract $3$ from both sides: $61.501 \approx 5t$.
Divide by $5$ to solve for $t$: $t \approx 12.3002$.
The time required is approximately $12.3$ years.
Correct Option: D
▶️ Answer/Explanation
Substitute $t = 5$ into $M(t)$ to get $260,000 = \frac{540,000}{1 + 0.4e^{5k}}$.
Simplify the fraction: $1 + 0.4e^{5k} = \frac{540,000}{260,000} \approx 2.0769$.
Solve for the exponential term: $0.4e^{5k} \approx 1.0769$, so $e^{5k} \approx 2.6923$.
Find the growth constant: $k = \frac{\ln(2.6923)}{5} \approx 0.19808$.
Substitute $t = 15$ and $k$ into the original function: $M(15) = \frac{540,000}{1 + 0.4e^{15(0.19808)}}$.
Calculate the denominator: $1 + 0.4(e^{5k})^3 = 1 + 0.4(2.6923)^3 \approx 8.8034$.
Final calculation: $M(15) = \frac{540,000}{8.8034} \approx 61,340$ (rounding gives $61,321$).
The correct option is C.
▶️ Answer/Explanation
The correct option is B.
A residual is calculated as $\text{Residual} = \text{Observed } y – \text{Predicted } \hat{y}$.
Since point $P$ has a negative residual of $-0.0257$, it must lie below the regression line.
A negative residual occurs when the $\text{Observed } y < \text{Predicted } \hat{y}$.
This indicates that the model’s predicted value is higher than the actual data point.
Therefore, the regression model provides an overestimate at this specific point.
Option A is incorrect because a lack of pattern actually suggests a linear model is appropriate.
▶️ Answer/Explanation
The given regression equation is $\ln(y) = 0.0568x + 5.226$.
To solve for $y$, exponentiate both sides using base $e$: $y = e^{0.0568x + 5.226}$.
Apply the exponent rule $e^{a+b} = e^{a} \cdot e^{b}$ to get $y = e^{5.226} \cdot e^{0.0568x}$.
Calculate the constant term: $e^{5.226} \approx 186.047$.
Calculate the base of the exponent: $e^{0.0568} \approx 1.0584$.
Substitute these values back to get $y = 186.047(1.0584)^{x}$.
Therefore, the correct option is D.
▶️ Answer/Explanation
The correct option is A.
In a semi-log plot, the vertical axis represents $\log(P)$ while the horizontal axis remains $t$.
Calculating the values: $\log(78) \approx 1.89$, $\log(55) \approx 1.74$, $\log(20) \approx 1.30$, $\log(9.6) \approx 0.98$, and $\log(5.4) \approx 0.73$.
The horizontal axis (time) must range from $1$ to $5$ days.
The vertical axis must range from approximately $0.7$ to $1.9$.
Graph A correctly displays the downward trend of $\log(P)$ over the time interval $t \in [1, 5]$.
Graph B is incorrect because its horizontal axis represents log values instead of time.
Graph C uses a linear scale for the vertical axis, and Graph D shows an incorrect upward trend.
▶️ Answer/Explanation
The correct option is B.
By performing logarithmic regression on the data points, we find the coefficients $a \approx 5.4294$ and $b \approx -3.13109$.
The regression equation is $\hat{y} = 5.4294 \ln(x) – 3.13109$.
To find the residual at $x = 5$, first calculate the predicted value: $\hat{y} = 5.4294 \ln(5) – 3.13109 \approx 5.6072$.
The observed value $y$ at $x = 5$ is $5.5$.
The residual is calculated as $e = y – \hat{y} = 5.5 – 5.6072 = -0.1072$.
Therefore, option B matches both the regression equation and the calculated residual.
▶️ Answer/Explanation
(A)
The value of \(c\) is determined by the horizontal asymptote of the exponential function. Looking at the graph, as \(x \to \infty\), the function \(g(x)\) approaches the line \(y = -1\).
Therefore, \(c = -1\).
(B)
We have the equation \(g(x) = a(b)^x – 1\). We can use the given points \((0, 1)\) and \((-1, 3)\) to solve for \(a\) and \(b\).
First, substitute \((0, 1)\):
\(1 = a(b)^0 – 1 \Rightarrow 1 = a(1) – 1 \Rightarrow a = 2\).
Now the equation is \(g(x) = 2(b)^x – 1\). Substitute \((-1, 3)\):
\(3 = 2(b)^{-1} – 1 \Rightarrow 4 = \frac{2}{b} \Rightarrow 4b = 2 \Rightarrow b = \frac{1}{2}\).
So, \(a = 2\) and \(b = \frac{1}{2}\).
(C)
To find the inverse function, we start with \(y = 2(2)^{-x} – 1\) (since \(b = 1/2 = 2^{-1}\)) and swap \(x\) and \(y\):
\(x = 2(2)^{-y} – 1\).
Isolate the exponential term:
\(x + 1 = 2(2)^{-y} \Rightarrow x + 1 = 2^{1-y}\).
Convert to logarithmic form:
\(\log_2(x+1) = 1 – y\).
Solve for \(y\):
\(y = 1 – \log_2(x+1)\).
Thus, \(g^{-1}(x) = 1 – \log_2(x+1)\).
(D)
The domain of the inverse function corresponds to the range of the original function, and the range of the inverse corresponds to the domain of the original.
Domain of \(g^{-1}(x)\): The range of \(g(x)\) is \((-1, \infty)\), so Domain \(g^{-1}(x) = (-1, \infty)\).
Range of \(g^{-1}(x)\): The domain of \(g(x)\) is \((-\infty, \infty)\), so Range \(g^{-1}(x) = (-\infty, \infty)\).
▶️ Answer/Explanation
A.
An exponential regression model best represents the data.
This is because the output values (bacteria count) decrease proportionally rather than by a constant amount.
Additionally, the residuals of a linear fit would show a pattern, whereas they are randomly scattered for an exponential fit.
The model found is: \( \hat{y} = 279.401(0.7522)^x \).
B.
To linearize the data, we transform the dependent variable \( P \) using the natural logarithm, \( \ln(P) \).
Performing a linear regression on the transformed data points \( (t, \ln P) \) yields a linear equation.
The resulting linearization equation is: \( \ln P = -0.2847x + 5.6326 \).
C.
The slope \( m = -0.2847 \) represents the rate of change of the natural log of the bacteria population over time.
In context, it means the natural log of the bacteria count decreases by \( 0.2847 \) for every \( 1 \) day increase.
This implies the bacteria are decaying at a continuous rate determined by this slope.
D.
Using the exponential regression model from Part A: \( \hat{y}(14) = 279.401(0.7522)^{14} \).
Calculating the value: \( \hat{y} \approx 279.401(0.01857) \).
The predicted population on day 14 is approximately \( 5.189 \) thousand bacteria.
E.
Starting with the linearized equation: \( \ln y = -0.2847x + 5.6326 \).
Exponentiate both sides to isolate \( y \): \( y = e^{-0.2847x + 5.6326} \).
Using exponent rules \( e^{a+b} = e^a \cdot e^b \): \( y = (e^{-0.2847})^x \cdot e^{5.6326} \).
Solving for the constants gives: \( y \approx 279.388(0.7522)^x \).