▶️ Answer/Explanation
The correct option is (B).
Apply the power rule: $n \ln x = \ln x^n$, giving $\ln w^2 – \ln y – \ln z^3$.
Factor out the negative sign: $\ln w^2 – (\ln y + \ln z^3)$.
Apply the product rule: $\ln y + \ln z^3 = \ln(y \cdot z^3)$.
Apply the quotient rule: $\ln w^2 – \ln(y \cdot z^3) = \ln \left( \frac{w^2}{y \cdot z^3} \right)$.
This matches the simplified expression shown in the handwritten note.
Therefore, the expression is condensed into a single logarithm.
▶️ Answer/Explanation
To find the intersection, set $f(x) = g(x)$, which gives $\log_{10}(x+3) – \log_{10}(x) = 1$.
Use the quotient rule for logarithms: $\log_{10}\left(\frac{x+3}{x}\right) = 1$.
Rewrite the logarithmic equation in exponential form: $\frac{x+3}{x} = 10^1$.
Multiply both sides by $x$ to get $x + 3 = 10x$.
Subtract $x$ from both sides: $3 = 9x$.
Solve for $x$ by dividing: $x = \frac{3}{9} = \frac{1}{3}$.
Since $x = \frac{1}{3}$ is positive, it satisfies the domain $x > 0$.
The correct option is (B).
▶️ Answer/Explanation
Substitute the given expressions: $\ln(2x + 1) = 2\ln 3$.
Use the power property of logarithms: $\ln(2x + 1) = \ln(3^2)$.
Simplify the right side: $\ln(2x + 1) = \ln 9$.
Remove the logarithms: $2x + 1 = 9$.
Solve for $x$: $2x = 8$.
Final result: $x = 4$.
The correct option is (D).
▶️ Answer/Explanation
Set the functions equal: $\log_{2}(3x – 1) – \log_{2}(x + 1) = \log_{2}(2)$.
Use the quotient rule for logarithms: $\log_{2}\left(\frac{3x – 1}{x + 1}\right) = \log_{2}(2)$.
Equate the arguments: $\frac{3x – 1}{x + 1} = 2$.
Multiply both sides by $(x + 1)$: $3x – 1 = 2(x + 1)$.
Expand and simplify: $3x – 1 = 2x + 2$.
Solve for $x$: $x = 3$.
Check domain: $3(3)-1 > 0$ and $3+1 > 0$, so $x = 3$ is valid.
The correct option is (C).
▶️ Answer/Explanation
To find the intersection, set the functions equal: $f(x) = g(x)$.
Substitute the given expressions: $\log_{3} x – \log_{3} 4 = 2$.
Apply the quotient rule for logarithms: $\log_{3} \left( \frac{x}{4} \right) = 2$.
Rewrite the logarithmic equation in exponential form: $\frac{x}{4} = 3^{2}$.
Simplify the exponent: $\frac{x}{4} = 9$.
Multiply both sides by $4$ to solve for $x$: $x = 36$.
The $x$-coordinate of the point of intersection is 36.
Therefore, the correct option is (D).
▶️ Answer/Explanation
Set the functions equal to find the intersection: $\log_{10}(x+3) + \log_{10}(x-5) = \log_{10}(3x-1)$.
Apply the product rule for logarithms: $\log_{10}((x+3)(x-5)) = \log_{10}(3x-1)$.
Equate the arguments: $(x+3)(x-5) = 3x-1$.
Expand and simplify to a quadratic: $x^2 – 2x – 15 = 3x – 1$, which becomes $x^2 – 5x – 14 = 0$.
Factor the quadratic equation: $(x-7)(x+2) = 0$, yielding potential solutions $x = 7$ and $x = -2$.
Check the domain: for $f(x)$, $x+3 > 0$ and $x-5 > 0$, requiring $x > 5$.
Since $x = -2$ is not in the domain ($x-5$ would be negative), only $x = 7$ is valid.
Correct Option: (D) $7$ only
▶️ Answer/Explanation
Set $f(x) = g(x)$, so $\ln((2x+3)(x-1)) = \ln(x^2 + 6x – 9)$.
Simplify the arguments: $2x^2 + x – 3 = x^2 + 6x – 9$.
Rearrange into a quadratic equation: $x^2 – 5x + 6 = 0$.
Factor the quadratic to find potential solutions: $(x-2)(x-3) = 0$, giving $x = 2$ and $x = 3$.
Check domain constraints: $2x+3 > 0$, $x-1 > 0$, and $x^2+6x-9 > 0$.
For $x=2$: $2(2)+3=7$, $2-1=1$, and $4+12-9=7$ (all positive).
For $x=3$: $2(3)+3=9$, $3-1=2$, and $9+18-9=18$ (all positive).
Both $x=2$ and $x=3$ are valid points of intersection.
The correct option is (A).
▶️ Answer/Explanation
The correct option is (D).
Given: $\log_{2}(x^{2}) – \log_{2} 4 = 3$
Use quotient rule: $\log_{2}\left(\frac{x^{2}}{4}\right) = 3$
Convert to exponential form: $\frac{x^{2}}{4} = 2^{3}$
Simplify the power: $\frac{x^{2}}{4} = 8$
Multiply both sides by 4: $x^{2} = 32$
Solve for $x$: $x = \pm\sqrt{32}$
Simplify the radical: $x = \pm 4\sqrt{2}$
▶️ Answer/Explanation
The correct option is (B).
Apply the product property of logarithms: \(\ln(a) + \ln(b) = \ln(ab)\).
Combine the terms: \(\ln(3x \cdot 2) = 4\), which simplifies to \(\ln(6x) = 4\).
Convert the natural logarithm to exponential form: \(e^4 = 6x\).
Isolate \(x\) by dividing both sides by 6: \(x = \frac{e^4}{6}\).
Verify the domain: since \(x > 0\), the solution is valid.
▶️ Answer/Explanation
Express the constant as a logarithm: $2 = \log_{7}(7^2) = \log_{7}(49)$.
Apply the product rule: $\log_{7}(x-1) + \log_{7}(49) = \log_{7}(49(x-1))$.
Set the arguments equal: $49(x-1) = 4x + 6$.
Expand and simplify: $49x – 49 = 4x + 6$.
Solve for $x$: $45x = 55$, which reduces to $x = \frac{11}{9}$.
Check domain constraints: $x-1 > 0 \implies x > 1$ and $4x+6 > 0 \implies x > -1.5$.
Since $\frac{11}{9} \approx 1.22$ satisfies $x > 1$, the solution is valid.
The correct option is (C).
▶️ Answer/Explanation
The correct answer is (A).
Rewrite the base $4$ as $2^2$ to get $(2^2)^{2x+6} = 2^{3x}$.
Apply the power of a power rule: $2^{2(2x+6)} = 2^{3x}$.
Since the bases are equal, set the exponents equal: $2(2x+6) = 3x$.
Distribute the $2$: $4x + 12 = 3x$.
Subtract $3x$ from both sides: $x + 12 = 0$.
Solve for $x$: $x = -12$.
▶️ Answer/Explanation
The point of intersection occurs where $h(x) = k(x)$.
Set the equations equal: $9^{(3x-1)} = 81^{(x+3)}$.
Express $81$ as a power of $9$, specifically $81 = 9^2$.
Substitute to get $9^{(3x-1)} = (9^2)^{(x+3)}$.
Apply the power rule $(a^m)^n = a^{mn}$ to get $9^{(3x-1)} = 9^{2(x+3)}$.
Equate the exponents: $3x – 1 = 2(x + 3)$.
Simplify and solve: $3x – 1 = 2x + 6$.
Subtract $2x$ and add $1$ to both sides to find $x = 7$.
The correct option is (D).
▶️ Answer/Explanation
To find the intersection, set $m(x) = p(x)$, giving $4^{(x^2 – 2x)} = \left( \frac{1}{16} \right)^{x-2}$.
Express both sides with base $4$: $4^{(x^2 – 2x)} = (4^{-2})^{x-2}$.
Simplify the right side power: $4^{(x^2 – 2x)} = 4^{-2x + 4}$.
Equate the exponents: $x^2 – 2x = -2x + 4$.
Add $2x$ to both sides to simplify: $x^2 = 4$.
Solve for $x$: $x = 2$ and $x = -2$.
The correct option is (B).
▶️ Answer/Explanation
Detailed solution:
To find the intersection, set the functions equal: $h(x) = k(x)$.
Substitute the expressions: $3^x = 9^{x-1}$.
Rewrite $9$ as a power of $3$: $3^x = (3^2)^{x-1}$.
Apply the power of a power rule: $3^x = 3^{2(x-1)}$.
Equate the exponents since the bases are the same: $x = 2(x-1)$.
Distribute the $2$: $x = 2x – 2$.
Subtract $x$ and add $2$ to solve: $x = 2$.
▶️ Answer/Explanation
To find the intersection, set the functions equal: $g(x) = h(x)$.
This gives the equation: $4^x = \left(\frac{1}{2}\right)^{x+3}$.
Express both sides with base $2$: $(2^2)^x = (2^{-1})^{x+3}$.
Simplify the exponents: $2^{2x} = 2^{-(x+3)}$.
Set the exponents equal to each other: $2x = -x – 3$.
Add $x$ to both sides: $3x = -3$.
Divide by $3$: $x = -1$.
The correct option is (C).
▶️ Answer/Explanation
Set the function equal to the target value: $3 + 2e^{-x} = 11$.
Subtract $3$ from both sides to isolate the exponential term: $2e^{-x} = 8$.
Divide both sides by $2$ to get $e^{-x} = 4$.
Take the natural logarithm ($\ln$) of both sides: $-x = \ln 4$.
Multiply by $-1$ to solve for $x$: $x = -\ln 4$.
The correct option is (C).
▶️ Answer/Explanation
The correct option is (B).
Given the function $g(x) = 3^{x+1}$, we set the equation to $3^{x+1} = 27$.
Express $27$ as a power of $3$, which is $3^3$.
Rewrite the equation as $3^{x+1} = 3^3$.
Since the bases are equal, set the exponents equal: $x + 1 = 3$.
Subtract $1$ from both sides to solve for $x$.
The result is $x = 2$.
▶️ Answer/Explanation
Set the function equal to the target value: $2^{3x-1} = \frac{1}{16}$.
Express $\frac{1}{16}$ as a power of 2: $\frac{1}{16} = \frac{1}{2^4} = 2^{-4}$.
Equate the exponents since the bases are the same: $3x – 1 = -4$.
Add 1 to both sides of the equation: $3x = -3$.
Divide both sides by 3 to solve for $x$: $x = -1$.
Therefore, the correct value is $x = -1$, which corresponds to option (B).
▶️ Answer/Explanation
Set the function equal to the target value: $3 \cdot 8^x = 12$.
Divide both sides by $3$ to isolate the exponential term: $8^x = 4$.
Express both $8$ and $4$ as powers of the common base $2$: $(2^3)^x = 2^2$.
Apply the power of a power rule: $2^{3x} = 2^2$.
Equate the exponents since the bases are the same: $3x = 2$.
Solve for $x$ by dividing by $3$: $x = \frac{2}{3}$.
The correct option is (D).
▶️ Answer/Explanation
Express $g(x)$ with base $2$: $g(x) = (2^2)^{x+1} = 2^{2x+2}$.
Set up the equation for $f(x) = 1$: $\frac{2^{2x+2}}{2^{3x}} = 1$.
Apply the quotient rule for exponents: $2^{(2x+2) – 3x} = 2^0$.
Simplify the exponent: $2^{-x+2} = 2^0$.
Equate the exponents: $-x + 2 = 0$.
Solve for $x$: $x = 2$.
The correct option is (D).
▶️ Answer/Explanation
The correct option is (B).
Substitute $f(x)$ into $g(x)$ to find $h(x) = \log_{3}(9^{x-1})$.
Rewrite $9$ as $3^2$, giving $h(x) = \log_{3}((3^2)^{x-1}) = \log_{3}(3^{2x-2})$.
Simplify using log properties to get $h(x) = 2x – 2$.
Set up the inequality $2x – 2 > 3$.
Add $2$ to both sides to get $2x > 5$.
Divide by $2$ to find $x > \frac{5}{2}$, which is the interval $\left( \frac{5}{2}, \infty \right)$.
▶️ 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).
▶️ Answer/Explanation
The correct option is (A).
The graphed points show that as \( x \) increases linearly, \( y \) increases by a constant ratio (doubling).
This identifies the function \( f \) as an exponential relationship of the form \( y = a \cdot b^x \).
On a semi-log plot, the vertical axis represents \( \log(y) \) instead of \( y \).
Applying a logarithm to both sides gives \( \log(y) = \log(a) + x \log(b) \).
This equation follows the linear form \( Y = mx + c \), where \( Y = \log(y) \).
Thus, exponential data always appears as a straight line on a semi-logarithmic scale.
▶️ Answer/Explanation
The correct answer is (A).
On a semi-log plot, a straight line represents an exponential function of the form $f(x) = ab^x$.
Observing the y-intercept at $x = 0$, the value is slightly below $1000$, which matches the initial value $a = 800$.
As $x$ increases, the $y$ values decrease, indicating an exponential decay where the base $b$ must be less than $1$.
At $x = 1$, the point is at $400$, which satisfies $800 \left( \frac{1}{2} \right)^1 = 400$.
At $x = 3$, the point is at $100$, which satisfies $800 \left( \frac{1}{2} \right)^3 = 800 \cdot \frac{1}{8} = 100$.
Option (C) is linear and (D) is logarithmic, neither of which produces a straight line on a semi-log scale.
▶️ Answer/Explanation
The correct answer is (A) $-\ln 5$.
In a semi-log plot with a natural log scale, the vertical coordinate is $y = \ln(g(x))$.
Substitute the function: $y = \ln\left( 6 \cdot \left( \frac{1}{5} \right)^x \right)$.
Use log properties: $y = \ln 6 + \ln\left( \left( \frac{1}{5} \right)^x \right)$.
Bring down the exponent: $y = \ln 6 + x \ln\left( \frac{1}{5} \right)$.
Apply the division rule for logs: $y = \ln 6 + x(\ln 1 – \ln 5)$.
Since $\ln 1 = 0$, the linear equation becomes $y = (-\ln 5)x + \ln 6$.
The slope of this linear relationship is the coefficient of $x$, which is $-\ln 5$.
▶️ Answer/Explanation
The correct option is (D).
A semi-log plot uses a logarithmic scale for the vertical axis, where increments represent powers of $10$.
In Graph D, the y-axis labels are $1, 10, 100, 1000$, which indicates a $\log_{10}$ scale.
Plotting the data: at $x=1, f(x)=12$ is just above $10$.
At $x=3, f(x)=108$ is just above $100$.
At $x=5, f(x)=980$ is just below $1000$.
These points form a nearly straight line on the semi-log grid, characteristic of exponential growth.
▶️ Answer/Explanation
The correct answer is (A).
The values of \(g(x)\) decrease by a constant factor of approximately \(0.45\), indicating an exponential function.
In a semi-log plot, an exponential function \(g(x) = ab^x\) appears as a straight line.
The vertical axis in Graph A uses a base-\(10\) logarithmic scale spanning from \(10^0\) to \(10^3\).
Point \((1, 813)\) is correctly placed near the top grid line of \(1000\).
Point \((5, 33)\) is correctly placed between \(10\) and \(100\).
The linear alignment of points in Graph A confirms the relationship \(\log(g(x)) = \log(a) + x\log(b)\).
Options (B), (C), and (D) are incorrect due to improper scaling or the use of linear axes.
▶️ Answer/Explanation
The correct answer is (B).
Observe that as \(x\) increases by a constant amount of \(5\), the value of \(h(x)\) is multiplied by a constant factor of \(3\).
This constant ratio (\(6/2 = 3\), \(18/6 = 3\), etc.) indicates that \(h(x)\) is an exponential function.
An exponential function is defined by the general form \(y = ab^x\).
Taking the logarithm of both sides gives \(\log(y) = \log(a) + x\log(b)\).
This transformation shows a linear relationship between \(x\) and \(\log(y)\).
Therefore, on a semi-log plot with a logarithmic vertical axis, the data will appear as a straight line.
▶️ Answer/Explanation
The correct option is (A).
A straight line on a semi-log plot indicates an exponential relationship between $x$ and $y$.
The linear trend follows the form $\log_{10}(y) = mx + b$, which simplifies to $y = 10^{mx+b}$.
Since the slope $m$ is positive, the data represents exponential growth.
In a standard $xy$-plane, the rate of change of an exponential growth function increases as $x$ increases.
This increasing slope results in a graph that is concave up.
Logarithmic functions, by contrast, would appear linear on a log-log plot or semi-log $x$ plot.
▶️ Answer/Explanation
In a semi-log plot with a logarithmic vertical axis, the linear equation is written as $\log_{10}(f(x)) = y$.
Substitute the given linear model: $\log_{10}(f(x)) = (\log_{10} 2)x + \log_{10} 5$.
Apply the power property of logarithms: $\log_{10}(f(x)) = \log_{10}(2^x) + \log_{10} 5$.
Apply the product property of logarithms: $\log_{10}(f(x)) = \log_{10}(5 \cdot 2^x)$.
Take the antilogarithm of both sides: $f(x) = 5 \cdot 2^x$.
This matches option (B).
▶️ Answer/Explanation
The semi-log plot represents the relationship as $y = \ln(g(x))$.
Given the linear model: $y = \ln 4 – x(\ln 3)$.
Substitute $y$ with $\ln(g(x))$ to get: $\ln(g(x)) = \ln 4 – \ln(3^x)$.
Apply logarithm rules to combine terms: $\ln(g(x)) = \ln\left(\frac{4}{3^x}\right)$.
Exponentiate both sides to solve for $g(x)$: $g(x) = 4 \cdot \left(\frac{1}{3}\right)^x$.
Compare this to the standard form $g(x) = ab^x$.
The constant $b$ is identified as $\frac{1}{3}$.
Thus, the correct option is (D).
▶️ Answer/Explanation
The correct option is (C).
A straight line on a semi-log plot indicates an exponential relationship between variables.
Since the data points follow a linear trend on the \(\log\) scale, model \(P\) must be exponential.
The residual plot shows points randomly scattered around the zero line with no discernible pattern.
A random residual plot confirms that the chosen regression model is appropriate for the data.
Therefore, model \(P\) is exponential and the fit is considered appropriate.