Math 3 Quiz: Exponential Logarithmic Inverse Relationship

What this quiz covers

This quiz focuses on Exponential Logarithmic Inverse Relationship, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The calculation proves $n(x) = m^{-1}(x)$ because $m(n(2)) = 2$ shows the composition returns the original input
2. The calculation confirms inverse functions because $\log_5(25) = 2$ validates the exponential-logarithmic relationship at this point
3. The calculation shows the functions are not inverses because $m(n(2)) \neq n(m(2))$ due to asymmetric composition results
4. The calculation demonstrates $m(n(x)) = x$ for $x = 2$, providing partial evidence that the functions are inverses (correct answer)

Question 2

1. The functions are not inverses because $g(f(x)) = \ln((e^{2x})^2) = \ln(e^{4x}) = 4x \neq x$ for the composition test
2. The functions are not inverses because $f(g(x)) = e^{2\ln(x^2)} = e^{\ln((x^2)^2)} = x^4 \neq x$ for all values of $x$ (correct answer)
3. The functions are not inverses because their domains are incompatible: $f(x)$ accepts all real numbers while $g(x)$ requires $x \neq 0$
4. The functions are not inverses because $g(x) = \ln(x^2) = 2\ln|x|$ is not one-to-one due to the absolute value relationship

Question 3

1. The composition $f(f^{-1}(48)) = 48$ demonstrates the fundamental inverse property for this specific input value
2. The symmetry condition $f(2) = f^{-1}(2)$ shows that both functions yield identical outputs for identical inputs
3. The equation $f^{-1}(f(2)) = 2$ confirms that applying inverse operations returns the original input value (correct answer)
4. The relationship $f(2) \cdot f^{-1}(48) = 96$ satisfies the multiplicative inverse property for exponential functions

Question 4

1. The point $(25, 2)$ lies on $q(x)$, and $q(x)$ has a horizontal asymptote at $y = 25$ corresponding to the vertical asymptote of $p(x)$
2. The point $(2, 25)$ lies on $q(x)$, and the graphs of $p(x)$ and $q(x)$ intersect at $(25, 2)$ due to their inverse relationship
3. The point $(2, 25)$ lies on $q(x)$, and $q(x)$ passes through $(0, 1)$ while $p(x)$ has a vertical asymptote at $x = 0$ (correct answer)
4. The point $(-2, -25)$ lies on $q(x)$ by the inverse relationship, and both functions have the same rate of change at corresponding points

Question 5

1. Apply $h(x) = e^{x+2} - 3$ to undo the shifts; this preserves the inverse relationship because exponential and logarithmic transformations are commutative
2. Apply $h(x) = e^{x-2} + 3$ to reverse the transformations; this preserves the inverse relationship because the composition returns to the original domain restrictions
3. Apply $h(x) = e^{x-2} - 3$ to partially undo the shifts; this maintains the inverse relationship through the preservation of the fundamental exponential-logarithmic connection
4. Apply $h(x) = e^{x+2} - 3$ to reverse the transformations; this preserves the inverse relationship because applying inverse operations in reverse order restores the original function (correct answer)

Question 6

1. The intersection occurs at $(1, 1)$ because $3^1 = 1$ and $\log_3(1) = 1$ both equal $1$
2. The intersection occurs where $3^a = a$ and $\log_3(a) = a$, with $a \approx 1.17$ as the unique solution (correct answer)
3. Multiple intersections exist because inverse functions always intersect at several points along $y = x$
4. No intersection exists because exponential and logarithmic functions have different monotonicity properties

Question 7

1. $h(4) = 11$ because $h(4) = 2^{4-1} + 3 = 2^3 + 3 = 8 + 3 = 11$ (correct answer)
2. $h(4) = 19$ because $h(4) = 2^{4+1} + 3 = 2^5 + 3 = 32 + 3 = 35$ divided by something
3. $h(4) = 5$ because the inverse transformation shifts the input by $2$ units in the opposite direction
4. $h(4) = 7$ because $h(4) = 2^{4} + 3 = 16 + 3 = 19$ but adjusted for the domain shift

Question 8

1. $h^{-1}(5) = 4$ because $4^{x/2} = 8$ gives $x/2 = \log_4(8) = 2$, so $x = 4$
2. $h^{-1}(5) = 2$ because $\log_4(8) = 1$, then $x/2 = 1$ gives $x = 2$
3. $h^{-1}(5) = 6$ because $4^{x/2} = 8$ gives $\log_4(8) = 1.5$, then $x = 2(1.5) \times 2 = 6$
4. $h^{-1}(5) = 3$ because $4^{x/2} = 8$ gives $\log_4(8) = \log_4(4^{3/2}) = 3/2$, so $x = 3$ (correct answer)

Question 9

1. $(-4, \infty)$ because the range of $h(x)$ excludes values at and below $-4$ (correct answer)
2. $(-\infty, \infty)$ because exponential functions are defined for all real numbers
3. $[0, \infty)$ because logarithmic inverses require non-negative arguments for real outputs
4. $(0, \infty)$ because the inverse involves $\log_2(x+4)$ which requires $x > -4$

Question 10

1. $f^{-1}(x) = \log_3(x) - 1 + 2$ obtained by taking the logarithm base $3$ and adjusting for the shifts
2. $f^{-1}(x) = \log_3(x + 1) - 2$ obtained by applying inverse operations to both the horizontal and vertical shifts
3. $f^{-1}(x) = \log_3(x - 1) + 2$ obtained by reversing the transformations in the appropriate order (correct answer)
4. $f^{-1}(x) = 3^{x+2} - 1$ obtained by reversing the signs of both transformation parameters in the original function

Question 11

1. The point $(8, 3)$ lies on the graph of $g(x)$, and the graphs are reflections of each other across the line $y = x$ (correct answer)
2. The point $(3, 8)$ also lies on the graph of $g(x)$, and the graphs are reflections of each other across the $x$-axis
3. The point $(-3, -8)$ lies on the graph of $g(x)$, and the graphs are reflections of each other across the origin
4. The point $(8, 3)$ lies on the graph of $g(x)$, and the graphs are reflections of each other across the $y$-axis

Question 12

1. Domain: $(2, \infty)$; $f^{-1}(x)$ has a horizontal asymptote at $y = 1$ and increases without bound as $x$ increases
2. Domain: $(2, \infty)$; $f^{-1}(x)$ has a vertical asymptote at $x = 2$ and approaches $-\infty$ as $x$ approaches $2^+$ (correct answer)
3. Domain: $(1, \infty)$; $f^{-1}(x)$ has a horizontal asymptote at $y = 2$ and decreases as $x$ increases beyond $2$
4. Domain: $(2, \infty)$; $f^{-1}(x)$ has a horizontal asymptote at $y = 2$ and increases logarithmically as $x$ increases from $2$

Question 13

1. They are inverses because $h(0) = 0$ and $k(0) = 0$, showing that both functions pass through the origin as required for inverse exponential-logarithmic pairs
2. They are not inverses because $h(k(3)) = 3$ but $k(h(3)) \neq 3$, violating the symmetric property required for inverse functions
3. They are not inverses because $h(x)$ has range $(-1, \infty)$ while $k(x)$ has domain $(-1, \infty)$, but their compositions don't simplify to the identity function
4. They are inverses because substituting $y = 4^x - 1$ gives $x = \log_4(y + 1)$, and their graphs are reflections across $y = x$ (correct answer)

Question 14

1. Inverse functions always intersect at points where $f(a) = a$, and these points lie on both graphs simultaneously
2. The reflection property across $y = x$ requires that intersection points satisfy both $y = f(x)$ and $x = f(y)$ with equal coordinates
3. At intersection points, both $2^x = y$ and $\log_2(x) = y$ hold, which requires $x = y$ by the definition of inverse functions (correct answer)
4. The symmetry of inverse functions across $y = x$ means intersection points are equidistant from both axes, requiring equal coordinates

Question 15

1. $b^{2k} = 4$; this shows that applying the exponential function to twice the logarithm gives the square of the original argument
2. $b^{2k} = 4$; this demonstrates that $b^{\log_b(2^2)} = 2^2$ by the inverse property of exponential and logarithmic functions (correct answer)
3. $b^{2k} = 8$; this shows that the exponential function undoes the logarithmic operation when applied to doubled inputs
4. $b^{2k} = 2$; this demonstrates that logarithmic and exponential operations cancel when the coefficient equals the logarithmic result