5.4 Graphs and Properties of Logarithmic Functions

Introduction

In this section, you will: 1. Examine properties of logarithmic functions 2. Examine graphs of logarithmic functions 3. Explore the relationship between graphs of exponential and logarithmic functions > Context Pause > Why should you care about what logarithmic graphs look like? In the real world, logarithmic scales are everywhere. The Richter scale for earthquakes, the decibel scale for sound, and the pH scale in chemistry all use logarithms. When you understand the shape and behavior of a log graph, you can quickly interpret data displayed on these scales -- for instance, why a magnitude 7 earthquake is not "a little worse" than a magnitude 6, but roughly 10 times more powerful. ---

Building a Logarithmic Table from an Exponential Table

Recall from Section 5.1 that the exponential function $f(x) = 2^x$ produces this table of values: | $x$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | |-----|------|------|------|-----|-----|-----|-----| | $f(x)$ | $1/8$ | $1/4$ | $1/2$ | $1$ | $2$ | $4$ | $8$ | Since the logarithmic function is the inverse of the exponential, $g(x) = \log_2(x)$ simply swaps the inputs and outputs. This gives us: | $x$ | $1/8$ | $1/4$ | $1/2$ | $1$ | $2$ | $4$ | $8$ | |-----|-------|-------|-------|-----|-----|-----|-----| | $g(x)$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | > Insight Note > Notice what happened: we literally flipped the rows. Every output from the exponential function became an input for the logarithm, and vice versa. This "row-swap" trick works because logarithms and exponentials are inverse functions -- they undo each other. ---

Key Observations About Logarithmic Output

Looking at the second table, we can make four important observations: 1. As the input increases, the output increases. When $x$ gets bigger, $g(x) = \log_2(x)$ also gets bigger. 2. The output increases more and more slowly. Going from $x = 1$ to $x = 2$ increases the output by 1 (from 0 to 1), but going from $x = 4$ to $x = 8$ also only increases the output by 1 (from 2 to 3), even though the input jumped by 4 instead of 1. 3. The domain is only positive numbers. Since the exponential function $2^x$ only outputs positive values, the logarithm can only accept positive inputs. The domain of the log function is $(0, \infty)$. 4. The range is all real numbers. Since the exponential function can accept any real number as input, the logarithm can produce any real number as output. The range is $(-\infty, \infty)$. > Context Pause > This "slowing down" behavior is exactly why logarithmic scales are so useful. If you are measuring earthquake intensities that range from tiny tremors to catastrophic events, a logarithmic scale compresses the enormous range into manageable numbers. A log graph "stretches out" the small values and "compresses" the large ones. ---

The Graph of a Basic Logarithmic Function

When we plot $g(x) = \log_2(x)$ using the points from our table, we see a characteristic curve. As the input values $x$ approach zero from the right, the output plunges downward without bound. This tells us there is a vertical asymptote at $x = 0$ (the $y$-axis). In symbolic notation, we write: $$\text{As } x \to 0^{+}, \quad f(x) \to -\infty$$\text{As } x \to \infty, \quad f(x) \to \infty$$ In plain English: as $x$ gets closer and closer to zero (from the positive side), the output drops to negative infinity. As $x$ grows without bound, the output also grows -- but very slowly. ![](../source_files/AppliedFiniteMath-3ed-Current/images/aee466cc131cbf86bd53382d9e2fd9b1eb3fadb8a5bd45c03a2b5ee809961872.jpg) Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions, Chapter 4, CC BY-SA 3.0 (modified by Roberta Bloom) ---

Properties of Logarithmic Functions ($b > 1$)

Definition 5.4.1: Properties of Logarithmic Functions with Base $b > 1$ Source: Applied Finite Math 3e (Lippman & Rasmussen, modified by Bloom) For the function $g(x) = \log_b(x)$ where $b > 1$: - The graph has a horizontal intercept (also called $x$-intercept) at the point $(1, 0)$ - The line $x = 0$ (the $y$-axis) is a vertical asymptote; as $x \to 0^{+}$, $y \to -\infty$ - The graph is increasing (rises from left to right) - Domain: $x > 0$, or $(0, \infty)$ - Range: all real numbers, or $(-\infty, \infty)$ ![](../source_files/AppliedFiniteMath-3ed-Current/images/51f9f435b099f5f55d99003aa73d3446733698cb49c9ad60fd6a0433f59b6b87.jpg) > Insight Note > Every logarithmic function with base greater than 1 passes through the point $(1, 0)$, regardless of the base. This makes sense: $\log_b(1) = 0$ for any base $b$, because $b^0 = 1$ always. The point $(1, 0)$ is the "home base" for all log graphs with $b > 1$. ---

Properties of Logarithmic Functions ($0 < b < 1$)

Definition 5.4.2: Properties of Logarithmic Functions with Base $0 < b < 1$ Source: Applied Finite Math 3e (Lippman & Rasmussen, modified by Bloom) If the base $b$ is between 0 and 1 (i.e., $0 < b < 1$), the graph flips compared to the $b > 1$ case. This follows from the reciprocal base property: $$\log_{1/b}(C) = -\log_b(C)$$ For the function $g(x) = \log_b(x)$ where $0 < b < 1$: - The graph still has a horizontal intercept at $(1, 0)$ - The line $x = 0$ (the $y$-axis) is still a vertical asymptote; but now as $x \to 0^{+}$, $y \to +\infty$ - The graph is decreasing (falls from left to right) - Domain: $x > 0$, or $(0, \infty)$ - Range: all real numbers, or $(-\infty, \infty)$ ![](../source_files/AppliedFiniteMath-3ed-Current/images/a2be1d6ccd88b70605ff3256d26d67d3dd1519ad86cc8cfd6bd4b61ee5ea2c6e.jpg) > Context Pause > When would you see a log function with a base less than 1? It is less common, but it arises naturally when measuring quantities that decrease on a proportional scale. For instance, in some signal processing applications, attenuation (weakening) of a signal can be modeled this way. The key takeaway is: base > 1 means increasing, and base between 0 and 1 means decreasing. ---

Graphing Tips for Logarithmic Functions

When graphing any logarithmic function by hand, two points are particularly easy to find: - The graph always passes through $(1, 0)$ -- because $\log_b(1) = 0$ for any base - The graph always passes through $(b, 1)$ -- because $\log_b(b) = 1$ for any base > Insight Note > These two "anchor points" -- $(1, 0)$ and $(b, 1)$ -- give you an instant sketch strategy. Plot those two points, draw the vertical asymptote at $x = 0$, then connect the dots with the characteristic log curve. If $b > 1$, the curve rises; if $0 < b < 1$, it falls. ---

The Mirror Relationship: Exponential and Logarithmic Graphs

Now let's put the graphs of $y = b^x$ and $y = \log_b(x)$ on the same set of axes. Because these functions are inverses of each other, for every ordered pair $(h, k)$ on the graph of $y = b^x$, the reversed pair $(k, h)$ appears on the graph of $y = \log_b(x)$. In other words: - If the point $(h, k)$ is on the graph of $y = b^x$ - Then the point $(k, h)$ is on the graph of $y = \log_b(x)$ This means: - The domain of $y = b^x$ equals the range of $y = \log_b(x)$ - The range of $y = b^x$ equals the domain of $y = \log_b(x)$ Because of this coordinate-swapping behavior, the two graphs appear as reflections (mirror images) of each other across the diagonal line $y = x$. This is a general property of all inverse function pairs that you may recall from prerequisite algebra. ![](../source_files/AppliedFiniteMath-3ed-Current/images/e9861137d90ad4c9b4dce309edd503492d9b0bce7fb61232e35da5da7e59d259.jpg) > Insight Note > Here is a quick way to remember the mirror relationship: if you fold the coordinate plane along the line $y = x$ (the 45-degree diagonal), the exponential graph lands exactly on top of the logarithmic graph. Every point "reflects" across that line. ---

Side-by-Side Comparison Table

Definition 5.4.3: Comparison of Exponential and Logarithmic Functions ($b > 1$) Source: Applied Finite Math 3e (Lippman & Rasmussen, modified by Bloom) | Property | $y = b^x$, with $b > 1$ | $y = \log_b(x)$, with $b > 1$ | |----------|--------------------------|-------------------------------| | Domain | All real numbers $(-\infty, \infty)$ | All positive real numbers $(0, \infty)$ | | Range | All positive real numbers $(0, \infty)$ | All real numbers $(-\infty, \infty)$ | | Intercepts | $(0, 1)$ -- a $y$-intercept | $(1, 0)$ -- an $x$-intercept | | Asymptotes | Horizontal asymptote: $y = 0$ (the $x$-axis); as $x \to -\infty$, $y \to 0$ | Vertical asymptote: $x = 0$ (the $y$-axis); as $x \to 0^{+}$, $y \to -\infty$ | > Context Pause > Notice the beautiful symmetry in this table. Every property of the exponential "swaps" for the logarithm: domain trades with range, $y$-intercept trades with $x$-intercept, and horizontal asymptote trades with vertical asymptote. This is not a coincidence -- it is the direct consequence of the inverse relationship. Whenever you forget a property of log functions, you can derive it by "flipping" the corresponding property of the exponential. Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions, Chapter 4, CC BY-SA 3.0 (modified by Roberta Bloom) ---

Problem Set: Section 5.4

Problem Set 5.4

Source: Applied Finite Math 3e Questions 1 - 3: For each of the following functions: a. Sketch a reasonably accurate graph showing the shape of the function b. State the domain c. State the range d. State whether the graph has a vertical asymptote or a horizontal asymptote and write the equation of that asymptote e. Does the graph have an $x$-intercept or a $y$-intercept? Write the coordinates of the intercept. --- 1) $y = \ln x$ a. Sketch the graph below b. Domain: ___________________ c. Range: ___________________ d. Is the asymptote horizontal or vertical? Equation of the asymptote: ___________________ e. Coordinates of $x$-intercept or $y$-intercept: ___________________

Problem 1 Solution

The function $y = \ln x$ is the natural logarithm, which means $y = \log_e(x)$ where the base is $e \approx 2.718$. Since $e > 1$, this is a logarithmic function with base greater than 1, so we apply Definition 5.4.1. Part (a): Sketch We use two anchor points from the Graphing Tips section: - The graph passes through $(1, 0)$ because $\ln(1) = 0$ - The graph passes through $(e, 1) \approx (2.718, 1)$ because $\ln(e) = 1$ Additional points: $\ln(1/e) = -1$ gives $(0.368, -1)$; $\ln(e^2) = 2$ gives $(7.389, 2)$ The curve rises from left to right, with a vertical asymptote at $x = 0$, increasing but becoming progressively flatter. Part (b): Domain Step 1: The logarithm is only defined for positive inputs (we can only take the log of a positive number). $$\text{Domain: } (0, \infty), \text{ i.e., } x > 0$$ Part (c): Range Step 1: As $x \to 0^+$, $\ln x \to -\infty$, and as $x \to \infty$, $\ln x \to \infty$. The output can be any real number. $$\text{Range: } (-\infty, \infty)$$ Part (d): Asymptote Step 1: As $x$ approaches 0 from the right, $\ln x$ plunges to $-\infty$. The graph gets closer and closer to the $y$-axis but never touches it. The asymptote is vertical. $$\text{Equation of the asymptote: } x = 0$$ Part (e): Intercept Step 1: Check for $x$-intercept by setting $y = 0$: $$\ln x = 0 \implies x = e^0 = 1$$ So the $x$-intercept is at $(1, 0)$. Step 2: Check for $y$-intercept by setting $x = 0$: $\ln(0)$ is undefined, so there is no $y$-intercept. Answer: The graph has an $x$-intercept at $(1, 0)$. Verification: $\ln(1) = 0$ ✓. The point $(1, 0)$ lies on the curve, confirming the $x$-intercept. The domain $(0, \infty)$ excludes $x = 0$, confirming no $y\(-intercept. ✓

--- 2) \)y = \log x$ a. Sketch the graph below b. Domain: ___________________ c. Range: ___________________ d. Is the asymptote horizontal or vertical? Equation of the asymptote: ___________________ e. Coordinates of $x$-intercept or $y\(-intercept: ___________________

Problem 2 Solution

The function \)y = \log x$ is the common logarithm (base 10), meaning $y = \log_{10}(x)$. Since the base $10 > 1$, this is a logarithmic function with base greater than 1, so we apply Definition 5.4.1. Part (a): Sketch We use the two anchor points: - The graph passes through $(1, 0)$ because $\log(1) = 0$ - The graph passes through $(10, 1)$ because $\log(10) = 1$ Additional points: $\log(0.1) = -1$ gives $(0.1, -1)$; $\log(100) = 2$ gives $(100, 2)$ The curve rises from left to right with a vertical asymptote at $x = 0$, increasing but progressively flatter. It looks similar to $y = \ln x$ but grows even more slowly (since base 10 > base $e$). Part (b): Domain Step 1: The common logarithm is only defined for positive inputs. $$\text{Domain: } (0, \infty), \text{ i.e., } x > 0$$ Part (c): Range Step 1: As $x \to 0^+$, $\log x \to -\infty$, and as $x \to \infty$, $\log x \to \infty$. $$\text{Range: } (-\infty, \infty)$$ Part (d): Asymptote Step 1: As $x \to 0^+$, $\log x \to -\infty$. The graph approaches the $y$-axis but never reaches it. The asymptote is vertical. $$\text{Equation of the asymptote: } x = 0$$ Part (e): Intercept Step 1: Check for $x$-intercept by setting $y = 0$: $$\log x = 0 \implies x = 10^0 = 1$$ So the $x$-intercept is at $(1, 0)$. Step 2: Check for $y$-intercept by setting $x = 0$: $\log(0)$ is undefined, so there is no $y$-intercept. Answer: The graph has an $x$-intercept at $(1, 0)$. Verification: $\log(1) = \log_{10}(1) = 0$ ✓. The point $(1, 0)$ is on the curve. Since $x = 0$ is not in the domain, there is no $y\(-intercept. ✓

--- 3) \)y = \log_{0.8} x$ a. Sketch the graph below b. Domain: ___________________ c. Range: ___________________ d. Is the asymptote horizontal or vertical? Equation of the asymptote: ___________________ e. Coordinates of $x$-intercept or $y\(-intercept: ___________________

Problem 3 Solution

The function \)y = \log_{0.8}(x)$ has base $b = 0.8$. Since $0 < 0.8 < 1$, this is a logarithmic function with base between 0 and 1, so we apply Definition 5.4.2. The key difference from Problems 1 and 2 is that the graph will be decreasing rather than increasing. Part (a): Sketch We use the two anchor points: - The graph passes through $(1, 0)$ because $\log_{0.8}(1) = 0$ - The graph passes through $(0.8, 1)$ because $\log_{0.8}(0.8) = 1$ Additional point: $\log_{0.8}(0.64) = 2$ because $0.8^2 = 0.64$, giving $(0.64, 2)$. Also: $\log_{0.8}(1.25) = -1$ because $0.8^{-1} = 1.25$, giving $(1.25, -1)$. The curve falls from left to right (decreasing), with a vertical asymptote at $x = 0$. As $x \to 0^+$, $y \to +\infty$ (note: this is positive infinity, opposite to the $b > 1$ case). Part (b): Domain Step 1: Logarithms are only defined for positive inputs, regardless of the base. $$\text{Domain: } (0, \infty), \text{ i.e., } x > 0$$ Part (c): Range Step 1: As $x \to 0^+$, $\log_{0.8}(x) \to +\infty$, and as $x \to \infty$, $\log_{0.8}(x) \to -\infty$. The output spans all real numbers. $$\text{Range: } (-\infty, \infty)$$ Part (d): Asymptote Step 1: As $x \to 0^+$, $\log_{0.8}(x) \to +\infty$. The graph approaches the $y$-axis but never touches it. The asymptote is vertical. $$\text{Equation of the asymptote: } x = 0$$ Part (e): Intercept Step 1: Check for $x$-intercept by setting $y = 0$: $$\log_{0.8}(x) = 0 \implies x = 0.8^0 = 1$$ So the $x$-intercept is at $(1, 0)$. Step 2: Check for $y$-intercept by setting $x = 0$: $\log_{0.8}(0)$ is undefined, so there is no $y$-intercept. Answer: The graph has an $x$-intercept at $(1, 0)$. Verification: $\log_{0.8}(1) = 0$ because $0.8^0 = 1$ ✓. The point $(1, 0)$ is on the curve. Since $x = 0$ is not in the domain, there is no $y\(-intercept. ✓

--- Questions 4 - 5: For the pair of inverse functions \)y = e^x$ and $y = \ln x$: a. Sketch a reasonably accurate graph showing the shape of the function b. State the domain c. State the range d. State whether the graph has a vertical asymptote or a horizontal asymptote and write the equation of that asymptote e. Does the graph have an $x$-intercept or a $y$-intercept? Write the coordinates of the intercept. --- 4) $y = e^x$ b. Domain: ___________________ c. Range: ___________________ d. Is the asymptote horizontal or vertical? Equation of the asymptote: ___________________ e. Coordinates of $x$-intercept or $y\(-intercept: ___________________

Problem 4 Solution

The function \)y = e^x$ is an exponential function with base $e \approx 2.718 > 1$. This is the inverse of $y = \ln x$ (Problem 5). We use the exponential properties from the Side-by-Side Comparison Table (Definition 5.4.3). Part (a): Sketch Key points: - At $x = 0$: $y = e^0 = 1$, so the graph passes through $(0, 1)$ - At $x = 1$: $y = e^1 \approx 2.718$, giving $(1, 2.718)$ - At $x = -1$: $y = e^{-1} \approx 0.368$, giving $(-1, 0.368)$ - At $x = 2$: $y = e^2 \approx 7.389$, giving $(2, 7.389)$ The curve rises steeply from left to right. As $x \to -\infty$, the graph approaches the $x$-axis from above but never touches it. Part (b): Domain Step 1: We can raise $e$ to any real number power. There is no restriction on the exponent. $$\text{Domain: } (-\infty, \infty)$$ Part (c): Range Step 1: The exponential function $e^x$ is always positive: $e^x > 0$ for all $x$. As $x \to -\infty$, $e^x \to 0^+$ (approaches but never reaches 0). As $x \to \infty$, $e^x \to \infty$. $$\text{Range: } (0, \infty)$$ Part (d): Asymptote Step 1: As $x \to -\infty$, $e^x \to 0$. The graph approaches the $x$-axis but never touches it. The asymptote is horizontal. $$\text{Equation of the asymptote: } y = 0$$ Part (e): Intercept Step 1: Check for $y$-intercept by setting $x = 0$: $$y = e^0 = 1$$ So the $y$-intercept is at $(0, 1)$. Step 2: Check for $x$-intercept by setting $y = 0$: $e^x = 0$ has no solution (exponentials are always positive), so there is no $x$-intercept. Answer: The graph has a $y$-intercept at $(0, 1)$. Verification: $e^0 = 1$ ✓. The range $(0, \infty)$ does not include 0, confirming no $x$-intercept. The domain $(-\infty, \infty)$ and range $(0, \infty)\( match the comparison table in Definition 5.4.3. ✓

--- 5) \)y = \ln x$ b. Domain: ___________________ c. Range: ___________________ d. Is the asymptote horizontal or vertical? Equation of the asymptote: ___________________ e. Coordinates of $x$-intercept or $y\(-intercept: ___________________

Problem 5 Solution

The function \)y = \ln x = \log_e(x)$ is the inverse of $y = e^x$ (Problem 4). Since the base $e \approx 2.718 > 1$, this is a logarithmic function with base greater than 1. We apply Definition 5.4.1. Note: This is the same function as Problem 1. In the context of Questions 4-5, Problems 4 and 5 are meant to be analyzed as a pair of inverse functions, highlighting how the properties of $y = e^x$ and $y = \ln x$ mirror each other. Part (a): Sketch Key points (same as Problem 1): - $(1, 0)$: $\ln(1) = 0$ - $(e, 1) \approx (2.718, 1)$: $\ln(e) = 1$ - $(1/e, -1) \approx (0.368, -1)$: $\ln(1/e) = -1$ When graphed alongside $y = e^x$, the two curves are reflections of each other across the line $y = x$. Each point $(h, k)$ on $y = e^x$ corresponds to the point $(k, h)$ on $y = \ln x$. Part (b): Domain Step 1: The logarithm is only defined for positive inputs. This equals the range of the inverse function $y = e^x$, which is $(0, \infty)$. $$\text{Domain: } (0, \infty)$$ Part (c): Range Step 1: The range of $\ln x$ equals the domain of its inverse $y = e^x$, which is all real numbers. $$\text{Range: } (-\infty, \infty)$$ Part (d): Asymptote Step 1: As $x \to 0^+$, $\ln x \to -\infty$. The graph approaches the $y$-axis but never touches it. The asymptote is vertical. (Compare: the inverse $y = e^x$ has a horizontal asymptote $y = 0$.) $$\text{Equation of the asymptote: } x = 0$$ Part (e): Intercept Step 1: Check for $x$-intercept: $\ln x = 0 \implies x = 1$, so $x$-intercept at $(1, 0)$. Step 2: No $y$-intercept since $x = 0$ is outside the domain. (Compare: the inverse $y = e^x$ has a $y$-intercept at $(0, 1)$ -- notice the coordinates are swapped!) Answer: The graph has an $x$-intercept at $(1, 0)$. Verification: $\ln(1) = 0$ ✓. Comparing with Problem 4: the $y$-intercept of $e^x$ at $(0, 1)$ swaps to the $x$-intercept of $\ln x$ at $(1, 0)\(, confirming the inverse relationship. ✓

--- Questions 6-11: Match the graph with the function. Choose the function from the list below and write it on the line underneath the graph. Hint: To match the function and the graph, identify these properties: - Is the function increasing or decreasing? - Examine the asymptote - Determine the \)x$- or $y$-intercept $$y = 3(2^x) \quad y = 5(0.4^x) \quad y = \log_2(x) \quad y = \log_{1/2}(x) \quad y = 3e^{-0.6x} \quad y = 5e^{0.3x}$\( ![](../source_files/AppliedFiniteMath-3ed-Current/images/fc6150fb93c55a62ef4763eaa5eb94d56d7ae13aeaf6c0f7aa56830034de3370.jpg)

Problem 6 Solution

Step 1: Identify the graph type. The graph is decreasing and approaches the \)x$-axis as $x \to \infty$ (horizontal asymptote at $y = 0$). This means it is an exponential decay function, not a logarithmic function (logarithmic functions have vertical asymptotes, not horizontal ones). Step 2: Identify the $y$-intercept. The graph crosses the $y$-axis at the point $(0, 3)$, so the $y$-intercept is $3$. Step 3: Narrow down the candidates. From the function list, the decreasing exponential functions are: - $y = 5(0.4^x)$: $y$-intercept at $y = 5(0.4^0) = 5(1) = 5$ - $y = 3e^{-0.6x}$: $y$-intercept at $y = 3e^0 = 3(1) = 3$ Since the $y$-intercept is $3$, the function must be $y = 3e^{-0.6x}$. Step 4: Confirm the behavior. For $y = 3e^{-0.6x}$: the exponent $-0.6x$ is negative when $x > 0$, so the function decreases. As $x \to \infty$, $e^{-0.6x} \to 0$, so $y \to 0$ (horizontal asymptote at $y = 0$). As $x \to -\infty$, $e^{-0.6x} \to \infty$, so the graph rises steeply to the left. This matches the graph. Answer: $y = 3e^{-0.6x}$ Verification: At $x = 0$: $y = 3e^{0} = 3$ ✓. The graph is decreasing with a horizontal asymptote at $y = 0\( ✓.

![](../source_files/AppliedFiniteMath-3ed-Current/images/b9e5207b2a8f7afc280a67308dd7c0e38264e991fce4cfa502d0c85a76adaa4f.jpg)

Problem 7 Solution

Step 1: Identify the graph type. The graph is increasing and has a vertical asymptote at \)x = 0$ (the $y$-axis). The graph passes through approximately $(1, 0)$. This is a logarithmic function with base greater than 1. Step 2: Identify the $x$-intercept. The graph crosses the $x$-axis at $(1, 0)$, consistent with $\log_b(1) = 0$ for any base $b$. Step 3: Narrow down the candidates. From the function list, the increasing logarithmic function is: - $y = \log_2(x)$: increasing (base $2 > 1$), $x$-intercept at $(1, 0)$ The other logarithmic candidate, $y = \log_{1/2}(x)$, is decreasing (base $1/2 < 1$), so it does not match. Step 4: Confirm the behavior. For $y = \log_2(x)$: at $x = 2$, $y = \log_2(2) = 1$; at $x = 4$, $y = \log_2(4) = 2$. The graph should pass through $(2, 1)$ and $(4, 2)$, which is consistent with the increasing curve shown. The vertical asymptote at $x = 0$ also matches. Answer: $y = \log_2(x)$ Verification: At $x = 1$: $\log_2(1) = 0$ ✓. At $x = 2$: $\log_2(2) = 1$ ✓. The function is increasing with a vertical asymptote at $x = 0\( ✓.

![](../source_files/AppliedFiniteMath-3ed-Current/images/8fdae926c7ab303fbbc2281ac4824e20f32bb1806598dd595cc7506cc16d5dd4.jpg)

Problem 8 Solution

Step 1: Identify the graph type. The graph is increasing and approaches the \)x$-axis as $x \to -\infty$ (horizontal asymptote at $y = 0$). This is an exponential growth function. Step 2: Identify the $y$-intercept. The graph crosses the $y$-axis at the point $(0, 3)$, so the $y$-intercept is $3$. Step 3: Narrow down the candidates. From the function list, the increasing exponential functions are: - $y = 3(2^x)$: $y$-intercept at $y = 3(2^0) = 3(1) = 3$ - $y = 5e^{0.3x}$: $y$-intercept at $y = 5e^0 = 5(1) = 5$ Since the $y$-intercept is $3$, the function must be $y = 3(2^x)$. Step 4: Confirm the behavior. For $y = 3(2^x)$: the base $2 > 1$, so the function increases. As $x \to -\infty$, $2^x \to 0$, so $y \to 0$ (horizontal asymptote at $y = 0$). As $x \to \infty$, $2^x \to \infty$, so the graph rises steeply to the right. This matches the graph. Answer: $y = 3(2^x)$ Verification: At $x = 0$: $y = 3(2^0) = 3(1) = 3$ ✓. At $x = 1$: $y = 3(2^1) = 6$ ✓. The graph is increasing with a horizontal asymptote at $y = 0\( ✓.

![](../source_files/AppliedFiniteMath-3ed-Current/images/cbfc1ee4c49fb77b67df05f4d77c1ec7423f09d2ecc813b737c22dbcf91eb4fc.jpg)

Problem 9 Solution

Step 1: Identify the graph type. The graph is decreasing and has a vertical asymptote at \)x = 0$ (the $y$-axis). The graph passes through $(1, 0)$. This is a logarithmic function with base between 0 and 1. Step 2: Identify the $x$-intercept. The graph crosses the $x$-axis at $(1, 0)$, consistent with $\log_b(1) = 0$ for any base $b$. Step 3: Narrow down the candidates. From the function list, the decreasing logarithmic function is: - $y = \log_{1/2}(x)$: decreasing (base $1/2 < 1$), $x$-intercept at $(1, 0)$ The other logarithmic candidate, $y = \log_2(x)$, is increasing (already matched to Graph 7). Step 4: Confirm the behavior. For $y = \log_{1/2}(x)$: at $x = 2$, $y = \log_{1/2}(2) = \frac{\ln 2}{\ln(1/2)} = \frac{\ln 2}{-\ln 2} = -1$. At $x = 1/2$, $y = \log_{1/2}(1/2) = 1$. So the graph passes through $(2, -1)$ and $(0.5, 1)$. As $x \to 0^+$, $y \to +\infty$ (the graph shoots up near the $y$-axis), and as $x \to \infty$, $y \to -\infty$ (the graph descends). This matches the decreasing curve shown. Answer: $y = \log_{1/2}(x)$ Verification: At $x = 1$: $\log_{1/2}(1) = 0$ ✓. At $x = 2$: $\log_{1/2}(2) = -1$ ✓. The function is decreasing with a vertical asymptote at $x = 0\( ✓.

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Problem 10 Solution

Step 1: Identify the graph type. The graph is increasing and approaches the \)x$-axis as $x \to -\infty$ (horizontal asymptote at $y = 0$). This is an exponential growth function. Step 2: Identify the $y$-intercept. The graph crosses the $y$-axis at the point $(0, 5)$, so the $y$-intercept is $5$. Step 3: Narrow down the candidates. From the remaining functions, the increasing exponential with $y$-intercept $5$ is: - $y = 5e^{0.3x}$: $y$-intercept at $y = 5e^0 = 5(1) = 5$ ✓ (The other candidate with $y$-intercept $5$, $y = 5(0.4^x)$, is decreasing since base $0.4 < 1$.) Step 4: Confirm the behavior. For $y = 5e^{0.3x}$: the exponent $0.3x$ is positive when $x > 0$, so the function increases. As $x \to -\infty$, $e^{0.3x} \to 0$, so $y \to 0$ (horizontal asymptote). As $x \to \infty$, $e^{0.3x} \to \infty$, so the graph rises steeply to the right. This matches the graph. Answer: $y = 5e^{0.3x}$ Verification: At $x = 0$: $y = 5e^{0} = 5$ ✓. The graph is increasing with a horizontal asymptote at $y = 0\( ✓.

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Problem 11 Solution

Step 1: Identify the graph type. The graph is decreasing and approaches the \)x$-axis as $x \to \infty$ (horizontal asymptote at $y = 0$). This is an exponential decay function. Step 2: Identify the $y$-intercept. The graph crosses the $y$-axis at the point $(0, 5)$, so the $y$-intercept is $5$. Step 3: Narrow down the candidates. The only remaining function is: - $y = 5(0.4^x)$: $y$-intercept at $y = 5(0.4^0) = 5(1) = 5$ ✓ This is also the only decreasing function with a $y$-intercept of $5$. Step 4: Confirm the behavior. For $y = 5(0.4^x)$: the base $0.4 < 1$, so the function decreases. As $x \to \infty$, $0.4^x \to 0$, so $y \to 0$ (horizontal asymptote). As $x \to -\infty$, $0.4^x \to \infty$, so the graph rises steeply to the left. This matches the graph. Step 5: Cross-check all assignments. | Graph | Function | Key Features | |-------|----------|-------------| | 6 | $y = 3e^{-0.6x}$ | Decreasing exponential, $y$-intercept $= 3$ | | 7 | $y = \log_2(x)$ | Increasing logarithmic, $x$-intercept $= (1,0)$ | | 8 | $y = 3(2^x)$ | Increasing exponential, $y$-intercept $= 3$ | | 9 | $y = \log_{1/2}(x)$ | Decreasing logarithmic, $x$-intercept $= (1,0)$ | | 10 | $y = 5e^{0.3x}$ | Increasing exponential, $y$-intercept $= 5$ | | 11 | $y = 5(0.4^x)$ | Decreasing exponential, $y$-intercept $= 5$ | All 6 functions have been matched exactly once. ✓ Answer: $y = 5(0.4^x)$ Verification: At $x = 0$: $y = 5(0.4^0) = 5(1) = 5$ ✓. At $x = 1$: $y = 5(0.4) = 2$ ✓. The graph is decreasing with a horizontal asymptote at $y = 0$ ✓.