Exponential-Logarithmic Inverse Relationship - Math 3

1. The function $h(x) = 2^{x+3} - 4$ has an inverse function. What is the domain of $h^{-1}(x)$?
   Answer: $(-4, \infty)$ because the range of $h(x)$ excludes values at and below $-4$ The domain of $h^{-1}(x)$ equals the range of $h(x)$. Since $h(x) = 2^{x+3} - 4$, and $2^{x+3} > 0$ for all real $x$, we have $h(x) > -4$. The range of $h(x)$ is $(-4, \infty)$, so the domain of $h^{-1}(x)$ is $(-4, \infty)$. Choice B confuses domain with range. Choice C incorrectly assumes logarithmic functions need non-negative inputs. Choice D makes an error in the inverse calculation.
2. If the range of $f(x) = 3^{x-1} + 2$ is $(2, \infty)$, what must be true about the domain and behavior of its inverse function $f^{-1}(x)$?
   Answer: Domain: $(2, \infty)$; $f^{-1}(x)$ has a vertical asymptote at $x = 2$ and approaches $-\infty$ as $x$ approaches $2^+$ The range of f becomes the domain of f⁻¹, so domain is (2,∞). Since f(x) = $3^{x-1}$ + 2 has a horizontal asymptote at y = 2 (as x → -∞), its inverse f⁻¹(x) has a vertical asymptote at x = 2. As x approaches 2⁺, f⁻¹(x) approaches -∞ because the original function approaches y = 2 as x → -∞. Choice A incorrectly describes a horizontal asymptote for the inverse. Choice C has wrong domain and wrong monotonicity. Choice D incorrectly describes a horizontal asymptote rather than vertical.
3. Two functions are defined as $m(x) = \log_5(2x + 1)$ and $n(x) = \frac{5^x - 1}{2}$. A student verifies these are inverses by computing $m(n(2)) = m\left(\frac{5^2 - 1}{2}\right) = m(12) = \log_5(25) = 2$. What does this calculation actually demonstrate?
   Answer: The calculation demonstrates $m(n(x)) = x$ for $x = 2$, providing partial evidence that the functions are inverses When testing whether two functions are inverses, you need to verify that both $f(g(x)) = x$ and $g(f(x)) = x$ for all values in their domains. This requires comprehensive verification, not just checking a single point. The student's calculation is mathematically correct: $m(n(2)) = m\left(\frac{5^2-1}{2}\right) = m(12) = \log_5(25) = 2$. This shows that $m(n(2)) = 2$, meaning the composition returns the original input when $x = 2$. However, this single verification only provides partial evidence that the functions might be inverses. Option A is incorrect because proving inverse functions requires showing the composition works for all values in the domain, not just one specific point. Option B misses the point entirely—while $\log_5(25) = 2$ is correct, this doesn't validate the inverse relationship; it's just confirming basic logarithm properties. Option C is wrong because the calculation doesn't show the functions aren't inverses, and the reasoning about "asymmetric composition results" doesn't apply here. Option D correctly identifies what happened: the calculation demonstrates that $m(n(x)) = x$ specifically when $x = 2$, which provides partial evidence toward proving the functions are inverses. To fully prove they're inverses, you'd need to verify this composition identity holds for all valid $x$ values, and also check that $n(m(x)) = x$. Remember: One successful composition at a single point suggests inverse functions but doesn't prove it. Always verify the relationship holds across the entire domain.
4. The graphs of $y = 2^x$ and $y = \log_2(x)$ intersect at exactly two points. Which property of inverse functions explains why these intersection points must have equal $x$ and $y$ coordinates?
   Answer: At intersection points, both $2^x = y$ and $\log_2(x) = y$ hold, which requires $x = y$ by the definition of inverse functions At intersection points, we need both y = $2^x$ and y = log₂(x) to be satisfied simultaneously. This means $2^x$ = log₂(x). For this equation to hold, we need x = y because if y = $2^x$, then x = log₂(y) by the inverse relationship. Setting these equal: if y = $2^x$ and y = log₂(x), then $2^x$ = log₂(x), which only occurs when x = y. Choice A is vague about the mechanism. Choice B mentions reflection but doesn't explain the coordinate relationship. Choice D describes geometric symmetry but not the algebraic necessity.
5. If $f(x) = 2^x$ and $g(x) = \log_2(x)$, and the point $(3, 8)$ lies on the graph of $f(x)$, what can be concluded about the relationship between the graphs of $f$ and $g$?
   Answer: The point $(8, 3)$ lies on the graph of $g(x)$, and the graphs are reflections of each other across the line $y = x$ Since f and g are inverse functions, their graphs are reflections across y = x. If (3, 8) is on f(x), then (8, 3) must be on g(x). We can verify: f(3) = 2³ = 8, and g(8) = log₂(8) = 3. Choice B incorrectly states the same point lies on both graphs. Choice C uses incorrect coordinates and wrong line of reflection. Choice D has correct coordinates but wrong line of reflection.
6. Consider the transformation $g(x) = \log_2(x - 3) + 1$. If $h(x)$ is the inverse of $g(x)$, what is $h(4)$?
   Answer: $h(4) = 11$ because $h(4) = 2^{4-1} + 3 = 2^3 + 3 = 8 + 3 = 11$ To find the inverse of $g(x) = \log_2(x - 3) + 1$, start with $y = \log_2(x - 3) + 1$. Solving for $x$: $y - 1 = \log_2(x - 3)$, so $2^{y-1} = x - 3$, giving $x = 2^{y-1} + 3$. Therefore $h(x) = 2^{x-1} + 3$. Computing $h(4) = 2^{4-1} + 3 = 2^3 + 3 = 8 + 3 = 11$. Choice B makes computational errors. Choice C incorrectly assumes a linear transformation. Choice D has the wrong exponent and unclear reasoning.
7. A student claims that $f(x) = e^{2x}$ and $g(x) = \ln(x^2)$ are inverse functions because "exponential and logarithmic functions are always inverses." What is the best explanation for why this reasoning is incorrect?
   Answer: 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 $x$ To verify inverse functions, both $f(g(x)) = x$ and $g(f(x)) = x$ must hold. Computing $f(g(x)) = f(\ln(x^2)) = e^{2\ln(x^2)} = e^{\ln((x^2)^2)} = (x^2)^2 = x^4 \neq x$ (except when $x = \pm 1$). This shows they are not inverses. Choice A makes an error in the composition calculation. Choice C identifies a domain issue but doesn't address the fundamental composition failure. Choice D correctly notes that $g(x)$ is not one-to-one, but choice B more directly demonstrates the inverse failure through computation.
8. Consider the function $h(x) = 4^{x/2} - 3$. If we restrict its domain to ensure it has an inverse, what is $h^{-1}(5)$?
   Answer: $h^{-1}(5) = 3$ because $4^{x/2} = 8$ gives $\log_4(8) = \log_4(4^{3/2}) = 3/2$, so $x = 3$ When working with inverse functions, you need to solve for the input value that produces a given output. Here, you're looking for the value of $x$ such that $h(x) = 5$. To find $h^{-1}(5)$, set up the equation $h(x) = 5$: $4^{x/2} - 3 = 5$ Solving step by step: $4^{x/2} = 8$ Now you need to find what power of 4 equals 8. Since $8 = 2^3$ and $4 = 2^2$, you can rewrite this as: $4^{x/2} = 8 = 2^3 = (2^2)^{3/2} = 4^{3/2}$ Therefore: $\frac{x}{2} = \frac{3}{2}$, which gives $x = 3$. Answer D correctly identifies this relationship: $\log_4(8) = \log_4(4^{3/2}) = \frac{3}{2}$, so $x = 3$. Answer A makes an error in calculating $\log_4(8)$, incorrectly stating it equals 2 instead of $\frac{3}{2}$. Answer B compounds this by claiming $\log_4(8) = 1$, which would mean $4^1 = 8$ (false, since $4^1 = 4$). Answer C starts correctly but then performs an unnecessary and incorrect multiplication by 2. The key strategy here is recognizing that logarithms and exponents are inverse operations. When you see $a^y = b$, you can solve for $y$ using $y = \log_a(b)$. Practice expressing numbers as powers of the same base to make these calculations cleaner.
9. If the exponential function $f(x) = 3^{x-2} + 1$ is reflected across the line $y = x$ to produce its inverse, what is the equation of the resulting function?
   Answer: $f^{-1}(x) = \log_3(x - 1) + 2$ obtained by reversing the transformations in the appropriate order When you encounter inverse function problems with exponential functions, remember that finding the inverse means swapping x and y, then solving for y. This process reveals how transformations in the original function affect the inverse. Starting with $f(x) = 3^{x-2} + 1$, let's find the inverse systematically. First, replace $f(x)$ with $y$: $y = 3^{x-2} + 1$. Next, swap x and y: $x = 3^{y-2} + 1$. Now solve for y by isolating the exponential term: $x - 1 = 3^{y-2}$. Take the logarithm base 3 of both sides: $\log_3(x-1) = y-2$. Finally, solve for y: $y = \log_3(x-1) + 2$. Therefore, $f^{-1}(x) = \log_3(x-1) + 2$. Choice A incorrectly writes the logarithm as $\log_3(x) - 1 + 2$, which fails to properly handle the vertical shift when finding the inverse. Choice B gives $\log_3(x + 1) - 2$, which incorrectly processes both transformations—the vertical shift of +1 in the original becomes $x + 1$ inside the logarithm (wrong), and the horizontal shift becomes -2 instead of +2. Choice D maintains the exponential form rather than converting to logarithmic form, which isn't how inverse functions work for exponentials. Remember this pattern: when finding inverses of transformed exponential functions, vertical shifts in the original function become restrictions on the domain (affecting what's inside the logarithm), while horizontal shifts become vertical shifts in the inverse. Always work through the algebra step-by-step rather than trying to apply transformation rules directly.
10. The graphs of $y = 3^x$ and $y = \log_3(x)$ intersect at point $(a, a)$. Which statement about this intersection is correct?
    Answer: The intersection occurs where $3^a = a$ and $\log_3(a) = a$, with $a \approx 1.17$ as the unique solution For the graphs to intersect at $(a, a)$, we need both $3^a = a$ and $\log_3(a) = a$ to be satisfied simultaneously. Since these are inverse functions, if $(a, a)$ lies on both graphs, then $a$ is a fixed point where $f(a) = a$. This occurs when $3^a = a$, which has a unique solution approximately $a \approx 1.17$. Choice A is wrong because $3^1 = 3 \neq 1$ and $\log_3(1) = 0 \neq 1$. Choice C is incorrect because inverse functions typically intersect at most at a few points. Choice D is false since both functions are increasing.
11. If $\log_b(16) = 4$ and $\log_b(2) = k$, what is the value of $b^{2k}$, and how does this demonstrate the inverse relationship between exponential and logarithmic functions?
    Answer: $b^{2k} = 4$; this demonstrates that $b^{\log_b(2^2)} = 2^2$ by the inverse property of exponential and logarithmic functions From $\log_b$(16) = 4, we get b⁴ = 16, so b = 2. From $\log_b$(2) = k, we get log₂(2) = 1, so k = 1. Therefore $b^{2k}$ = 2² = 4. This demonstrates the inverse property: since k = $\log_b$(2), we have $b^k$ = 2, and $b^{2k}$ = ($b^k$)² = 2² = 4, showing that $b^{log_b(x²)}$ = x². Choice A gives the correct value but incorrect reasoning. Choice C calculates the wrong value. Choice D gives wrong value and reasoning.
12. The function $f(x) = a \cdot b^{cx + d}$ has the inverse $f^{-1}(x) = \frac{1}{c}\log_b\left(\frac{x}{a}\right) - \frac{d}{c}$. If $f(2) = 48$ and $f^{-1}(48) = 2$, which property confirms these functions are true inverses?
    Answer: The equation $f^{-1}(f(2)) = 2$ confirms that applying inverse operations returns the original input value The fundamental property of inverse functions is that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ for all $x$ in the appropriate domains. Given $f(2) = 48$, we should have $f^{-1}(f(2)) = f^{-1}(48) = 2$, which matches the given information. This confirms the inverse relationship. Choice A tests the wrong composition direction. Choice B incorrectly suggests inverse functions should give the same output for the same input. Choice D invents a non-existent multiplicative property.
13. The function $p(x) = \log_5(x)$ has the property that $p(25) = 2$. Using the inverse relationship between exponential and logarithmic functions, what can be determined about the graph of $q(x) = 5^x$ without calculating additional points?
    Answer: The point $(2, 25)$ lies on $q(x)$, and $q(x)$ passes through $(0, 1)$ while $p(x)$ has a vertical asymptote at $x = 0$ Since p(25) = 2 means log₅(25) = 2, the inverse relationship tells us that q(2) = 5² = 25, so (2, 25) lies on q(x). Additionally, q(x) = $5^x$ always passes through (0, 1) since 5⁰ = 1, and p(x) = log₅(x) has a vertical asymptote at x = 0 since logarithms are undefined for non-positive arguments. Choice A incorrectly swaps coordinates and describes wrong asymptote behavior. Choice B has correct coordinates but incorrectly claims the graphs intersect at (25, 2). Choice D uses impossible negative coordinates for exponential functions.
14. Consider the transformation that maps $f(x) = e^x$ to $g(x) = \ln(x + 3) - 2$. What transformation maps $g(x)$ back to a function equivalent to $f(x)$, and why does this preserve the inverse relationship?
    Answer: 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 When working with inverse function transformations, you need to understand how shifts affect the exponential-logarithmic relationship and how to reverse them properly. Starting with $f(x) = e^x$ and $g(x) = \ln(x + 3) - 2$, notice that $g(x)$ represents a transformation of $\ln(x)$ (which is the inverse of $e^x$) with a horizontal shift left by 3 units and a vertical shift down by 2 units. To find the transformation that maps $g(x)$ back to something equivalent to $f(x)$, you need to reverse these operations in the correct order. Since $g(x) = \ln(x + 3) - 2$, applying the exponential function gives us $e^{g(x)} = e^{\ln(x + 3) - 2} = e^{-2} \cdot (x + 3)$. To get back to $e^x$, we need $h(x) = e^{x+2} - 3$. This undoes the vertical shift by adding 2 to the exponent, then undoes the horizontal shift by subtracting 3. Choice A incorrectly states that exponential and logarithmic transformations are commutative, which isn't the fundamental reason here. Choice B uses $e^{x-2} + 3$, which applies the transformations in the wrong direction. Choice C uses $e^{x-2} - 3$, which compounds the shifts rather than reversing them. Choice D correctly identifies $h(x) = e^{x+2} - 3$ and explains that applying inverse operations in reverse order restores the original function, which is the key principle governing inverse transformations. Study tip: When reversing transformations between inverse functions, always apply the opposite operations in reverse order to maintain the fundamental relationship.
15. Two students are debating whether $h(x) = 4^x - 1$ and $k(x) = \log_4(x + 1)$ are inverse functions. Which analysis correctly resolves their debate using properties of inverse functions?
    Answer: They are inverses because substituting $y = 4^x - 1$ gives $x = \log_4(y + 1)$, and their graphs are reflections across $y = x$ When you encounter questions about inverse functions, remember that two functions are inverses if and only if their compositions equal the identity function: $f(g(x)) = x$ and $g(f(x)) = x$ for all valid inputs. To verify that $h(x) = 4^x - 1$ and $k(x) = \log_4(x + 1)$ are inverses, let's check their compositions. First, $h(k(x)) = h(\log_4(x + 1)) = 4^{\log_4(x + 1)} - 1 = (x + 1) - 1 = x$. Second, $k(h(x)) = k(4^x - 1) = \log_4((4^x - 1) + 1) = \log_4(4^x) = x$. Since both compositions yield $x$, these functions are indeed inverses. Answer A is wrong because passing through the origin isn't required for inverse functions. In fact, $h(0) = 4^0 - 1 = 0$ but $k(0) = \log_4(1) = 0$ is coincidental, not a defining property. Answer B contains a computational error. When you calculate $h(k(3)) = h(\log_4(4)) = h(1) = 4^1 - 1 = 3$, and $k(h(3)) = k(63) = \log_4(64) = 3$, both equal 3, supporting that they are inverses. Answer C correctly identifies that the range of $h(x)$ matches the domain of $k(x)$, which is necessary for inverse functions, but incorrectly claims their compositions don't work. Study tip: Always verify inverse functions by checking that both compositions $f(g(x)) = x$ and $g(f(x)) = x$. Domain and range matching is necessary but not sufficient.

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