Exponential Functions - IB Mathematics Analysis and Approaches

1. Find the range of the function $f(x) = 7 - 3 \cdot 2^{x+1}$.
   Answer: $y < 7$ When finding the range of exponential functions, you need to analyze how the function behaves as the input varies across all real numbers, paying special attention to transformations applied to the basic exponential form. Let's rewrite $f(x) = 7 - 3 \cdot 2^{x+1}$ to better see its structure. Since $2^{x+1} = 2 \cdot 2^x$, we have $f(x) = 7 - 6 \cdot 2^x$. Now consider what happens to $2^x$ as $x$ changes: as $x \to \infty$, we get $2^x \to \infty$, and as $x \to -\infty$, we get $2^x \to 0^+$ (approaches zero from the positive side). This means $6 \cdot 2^x$ ranges from values approaching 0 to $+\infty$. Therefore, $f(x) = 7 - 6 \cdot 2^x$ ranges from $7 - \infty = -\infty$ up to (but never reaching) $7 - 0 = 7$. The range is $y < 7$. Looking at the wrong answers: Choice A suggests $y > 7$, which incorrectly assumes the function increases without bound—this ignores the negative coefficient. Choice B claims $y < -3$, likely from incorrectly thinking there's a lower bound at $y = -3$. Choice C suggests $y > -3$, which might come from misunderstanding the vertical shift or thinking the exponential term is positive. Strategy tip: For exponential functions of the form $a + b \cdot c^x$, identify the horizontal asymptote ($y = a$) and determine whether the function approaches it from above or below based on the sign of $b$. When $b < 0$, the range excludes values above the asymptote.
2. The half-life of a radioactive isotope is 20 days. If a sample initially contains 100g of the isotope, which function $M(t)$ correctly models the mass remaining after $t$ days?
   Answer: $M(t) = 100(2)^{-t/20}$ When you encounter exponential decay problems like radioactive isotopes, you're dealing with a situation where a quantity decreases by half over regular time intervals. The key is setting up the function so that when you substitute the half-life value, you get exactly half the original amount. For radioactive decay, the general form is $M(t) = M_0 \cdot b^{-t/h}$ where $M_0$ is the initial mass, $b$ is the base, and $h$ is the half-life. Since we want the mass to halve when $t = 20$, we need $M(20) = 50$. Let's check answer D: $M(t) = 100(2)^{-t/20}$. When $t = 20$: $M(20) = 100(2)^{-20/20} = 100(2)^{-1} = 100 \cdot \frac{1}{2} = 50$. Perfect! Answer A uses base 20 with exponent $-t/2$, which would give massive decay rates that don't match the 20-day half-life. Answer B has $0.5^{-t/20}$, but the negative exponent makes this represent growth, not decay—when $t = 20$, you get $0.5^{-1} = 2$, doubling the mass instead of halving it. Answer C uses $0.5^{20t}$, which creates extremely rapid decay—after just one day, you'd have virtually no isotope left. Remember: for half-life problems, always verify your function by substituting the half-life value for $t$. You should get exactly half the initial amount. This quick check will catch most errors and help you distinguish between exponential growth and decay functions.
3. The function $f(x) = 2^x$ is transformed to $g(x) = 2^{1-x} - 3$. Which statement accurately describes the transformation?
   Answer: Reflection in the y-axis, then a translation by $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$. When analyzing transformations of exponential functions, you need to carefully track how each algebraic change affects the graph's position and orientation. Let's break down $g(x) = 2^{1-x} - 3$ step by step. First, rewrite the exponent: $2^{1-x} = 2^1 \cdot 2^{-x} = 2 \cdot 2^{-x}$. Since $2^{-x} = (2^x)^{-1} = \frac{1}{2^x}$, we can think of this as starting with $f(x) = 2^x$, then applying transformations. The transformation $2^x \to 2^{-x}$ represents a reflection in the y-axis (replacing $x$ with $-x$). Then $2^{-x} \to 2 \cdot 2^{-x}$ is a vertical stretch by factor 2, and finally $2 \cdot 2^{-x} \to 2 \cdot 2^{-x} - 3$ shifts the graph down 3 units. However, we can also view $2^{1-x}$ as $2^{-(x-1)}$, which shows a reflection in the y-axis followed by a horizontal shift right by 1 unit. Combined with the vertical shift down 3 units, this gives us the translation vector $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$. Option A incorrectly suggests reflection in the x-axis (which would make the function negative). Option B has the wrong horizontal direction. Option C suggests a leftward translation when the shift is rightward. Only option D correctly identifies the y-axis reflection and the translation $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$. Study tip: When you see $f(-x)$, think y-axis reflection. When you see $-f(x)$, think x-axis reflection. Always rewrite complex transformations in the form $a \cdot f(b(x-h)) + k$ to identify each component clearly.
4. The function $f(x) = 4 \cdot 5^{x-1} + 3$ is obtained from the graph of $g(x) = 5^x$ by a sequence of transformations. Which of the following describes a correct sequence?
   Answer: A vertical stretch by a factor of 4, followed by a translation by the vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$. The equation is $f(x) = 4 \cdot 5^{x-1} + 3$. Let's analyze the transformations from $g(x)=5^x$. The term $x-1$ in the exponent corresponds to a horizontal translation of 1 unit to the right. The multiplication by 4 corresponds to a vertical stretch by a factor of 4. The addition of 3 corresponds to a vertical translation of 3 units up. A horizontal translation of 1 unit right and a vertical translation of 3 units up can be represented by the translation vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$. The order of transformations matters. The vertical stretch must be applied to the function $5^{x-1}$ before the vertical translation is applied. Let's check option A: Start with $5^x$. Apply a vertical stretch by factor 4 to get $4 \cdot 5^x$. Then translate by $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ (1 right, 3 up) to get $4 \cdot 5^{x-1} + 3$. This is correct. Let's check option B: Start with $5^x$. Translate by $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ to get $5^{x-1} + 3$. Then apply a vertical stretch by factor 4 to get $4(5^{x-1} + 3) = 4 \cdot 5^{x-1} + 12$, which is incorrect.
5. Consider the function $f(x) = k \cdot 2^{-x/3}$. The graph of $f$ passes through the point $(6, 5)$. Find the value of $f(-3)$.
   Answer: 40 First, we must find the value of the constant $k$ using the fact that the graph passes through $(6, 5)$. This means $f(6)=5$. Substitute $x=6$ and $f(x)=5$ into the function's equation: $5 = k \cdot 2^{-6/3}$ $5 = k \cdot 2^{-2}$ $5 = k \cdot \frac{1}{2^2}$ $5 = k \cdot \frac{1}{4}$ $k = 5 \times 4 = 20$. Now that we have $k$, the function is $f(x) = 20 \cdot 2^{-x/3}$. Next, we need to find the value of $f(-3)$. Substitute $x=-3$ into the function: $f(-3) = 20 \cdot 2^{-(-3)/3}$ $f(-3) = 20 \cdot 2^{3/3}$ $f(-3) = 20 \cdot 2^{1}$ $f(-3) = 20 \cdot 2 = 40$.
6. The temperature $T$, in degrees Celsius, of a cooling object $t$ minutes after it is removed from a heat source is given by $T(t) = 20 + 64{\left(\frac{1}{2}\right)}^{t/5}$. What is the temperature of the object after 10 minutes?
   Answer: 36°C We need to find the value of $T(10)$. Substitute $t=10$ into the given function: $T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{10/5}$ First, simplify the exponent: $\frac{10}{5} = 2$ Now the equation becomes: $T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{2}$ Calculate the power: ${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$ Now substitute this back: $T(10) = 20 + 64\left(\frac{1}{4}\right)$ $T(10) = 20 + 16$ $T(10) = 36$ The temperature after 10 minutes is 36°C.
7. A population of bacteria is modelled by $P(t) = 100a^t$, where $t$ is the time in hours. After 2 hours, the population is 900. At what time $t$ will the population reach 8100?
   Answer: 4 hours First, we use the given information to find the growth factor $a$. We know $P(2) = 900$. Substituting into the model: $100a^2 = 900$ $a^2 = 9$ Since $a$ must be positive for population growth, $a = 3$. So the model is $P(t) = 100(3)^t$. Next, we need to find the time $t$ when the population reaches 8100. $100(3)^t = 8100$ $3^t = 81$ We can write 81 as a power of 3: $81 = 9^2 = (3^2)^2 = 3^4$. So, $3^t = 3^4$. Therefore, $t = 4$ hours.
8. The graph of $y=2^x$ is reflected in the y-axis and then translated 3 units to the left. What is the equation of the transformed graph?
   Answer: $y = 2^{-x-3}$ Let's apply the transformations step-by-step to the function $f(x) = 2^x$. 1. Reflection in the y-axis: This transformation replaces $x$ with $-x$. The new function is $g(x) = f(-x) = 2^{-x}$. 2. Translation 3 units to the left: This transformation replaces $x$ with $(x+3)$ in the current function, $g(x)$. The final function is $h(x) = g(x+3) = 2^{-(x+3)}$. 3. Simplify the exponent: $h(x) = 2^{-x-3}$. The equation of the transformed graph is $y = 2^{-x-3}$. Distractor B arises from applying the transformations in the wrong order: translating left first gives $2^{x+3}$, and then reflecting in the y-axis gives $2^{-x+3}$.
9. Find the inverse function, $f^{-1}(x)$, of $f(x) = e^{x-3} + 4$.
   Answer: $f^{-1}(x) = \ln(x-4) + 3$ To find the inverse function, we start with the equation $y = f(x)$, then swap $x$ and $y$, and finally solve for $y$. 1. Start with $y = e^{x-3} + 4$. 2. Swap $x$ and $y$: $x = e^{y-3} + 4$. 3. Solve for $y$. First, isolate the exponential term: $x - 4 = e^{y-3}$ 4. To bring the exponent down, take the natural logarithm of both sides: $\ln(x-4) = \ln(e^{y-3})$ $\ln(x-4) = y-3$ 5. Isolate $y$: $y = \ln(x-4) + 3$. Therefore, the inverse function is $f^{-1}(x) = \ln(x-4) + 3$.
10. The value of a car, $V$, in dollars, depreciates according to the model $V(t) = 30000(0.8)^t$, where $t$ is the number of years since it was purchased. After how many full years will the value of the car first be less than ($10000)?
    Answer: 5 We need to find the smallest integer $t$ for which $V(t) < 10000$. Set up the inequality: $30000(0.8)^t < 10000$ Divide both sides by 30000: $(0.8)^t < \frac{10000}{30000}$ $(0.8)^t < \frac{1}{3}$ This could be solved with logarithms, but for a multiple choice question, we can test the integer values of $t$. $t=1: (0.8)^1 = 0.8$. Is $0.8 < 1/3$? No, $0.8 > 0.333...$. $t=2: (0.8)^2 = 0.64$. Still greater than $1/3$. $t=3: (0.8)^3 = 0.512$. Still greater than $1/3$. $t=4: (0.8)^4 = 0.4096$. Still greater than $1/3$. $t=5: (0.8)^5 = 0.32768$. This is less than $1/3 \approx 0.333...$. So, the first full year when the value is less than ($10000) is $t=5$.
11. The function $f(x) = a^x + b$ has a horizontal asymptote at $y=5$ and its graph passes through the point $(2, 21)$. Find the value of $a$.
    Answer: 4 The horizontal asymptote of the function $f(x) = a^x + b$ is given by the line $y=b$. We are given that the horizontal asymptote is $y=5$, so $b=5$. The function is therefore $f(x) = a^x + 5$. The graph passes through the point $(2, 21)$, which means $f(2) = 21$. We can substitute these values into the function's equation: $a^2 + 5 = 21$ $a^2 = 21 - 5$ $a^2 = 16$ Since $a$ is the base of an exponential function, it must be positive (and not equal to 1). Taking the positive square root, we find $a=4$.
12. What is the y-intercept of the graph of $f(x) = 5(3^{x+1} - 2)$?
    Answer: 5 The y-intercept of a function occurs where the graph crosses the y-axis, which is at $x=0$. To find the y-intercept, we need to calculate $f(0)$. Substitute $x=0$ into the function's equation: $f(0) = 5(3^{0+1} - 2)$ $f(0) = 5(3^1 - 2)$ $f(0) = 5(3 - 2)$ $f(0) = 5(1)$ $f(0) = 5$. The y-intercept is at $(0, 5)$. The value is 5.
13. Given $f(x) = 3^{2x}$ and $g(x) = x+1$, find the solution to the equation $(f \circ g)(x) = \frac{1}{9}$.
    Answer: -2 First, we find the composite function $(f \circ g)(x) = f(g(x))$. Substitute $g(x)$ into $f(x)$: $f(g(x)) = 3^{2(g(x))} = 3^{2(x+1)}$. Distribute the 2 in the exponent: $f(g(x)) = 3^{2x+2}$. Now, set this equal to $\frac{1}{9}$: $3^{2x+2} = \frac{1}{9}$. To solve this, we need to express both sides with the same base. Since $9 = 3^2$, we have $\frac{1}{9} = \frac{1}{3^2} = 3^{-2}$. So the equation becomes: $3^{2x+2} = 3^{-2}$. Now we can equate the exponents: $2x+2 = -2$ $2x = -4$ $x = -2$.
14. A substance decays according to the model $M(t) = M_0 (0.85)^t$, where $t$ is in years. What is the annual percentage decay rate of the substance?
    Answer: 15% The general model for exponential decay is $M(t) = M_0 (1-r)^t$, where $r$ is the annual decay rate. The term $(1-r)$ is the decay factor. In the given model, $M(t) = M_0 (0.85)^t$, the decay factor is 0.85. So, we have $1-r = 0.85$. Solving for $r$: $r = 1 - 0.85 = 0.15$. To express this rate as a percentage, we multiply by 100: $0.15 \times 100\% = 15\%$. The annual percentage decay rate is 15%.
15. Find the smallest integer $x$ for which $3 \cdot 2^x > 300$.
    Answer: 7 First, we simplify the inequality: $3 \cdot 2^x > 300$ Divide both sides by 3: $2^x > 100$ We need to find the smallest integer $x$ that satisfies this condition. We can test powers of 2: $2^1 = 2$ $2^2 = 4$ $2^3 = 8$ $2^4 = 16$ $2^5 = 32$ $2^6 = 64$ (This is less than 100, so $x=6$ is not the solution). $2^7 = 128$ (This is greater than 100). Therefore, the smallest integer value of $x$ for which $2^x > 100$ is $x=7$.
16. What is the solution to the equation $4^{x} - 6 \cdot 2^{x} - 16 = 0$?
    Answer: $x = 3$ This equation is a quadratic in disguise. We can rewrite $4^x$ as $(2^2)^x = (2^x)^2$. Let $y = 2^x$. The equation becomes: $y^2 - 6y - 16 = 0$ We can factor this quadratic equation: $(y-8)(y+2) = 0$ This gives two possible solutions for $y$: $y=8$ or $y=-2$. Now we substitute back $y = 2^x$: Case 1: $2^x = 8$. Since $8 = 2^3$, we have $x=3$. Case 2: $2^x = -2$. The exponential function $2^x$ is always positive for real $x$, so this case yields no real solution. The only solution is $x=3$.
17. An exponential function of the form $f(x) = ka^x$ passes through the points $(2, 12)$ and $(3, 18)$. Find the value of $f(1)$.
    Answer: 8 To find the parameters of $f(x) = ka^x$, we set up a system of two equations using the given points: 1. $ka^2 = 12$ 2. $ka^3 = 18$ Divide the second equation by the first to eliminate $k$: $$\frac{ka^3}{ka^2} = \frac{18}{12}$$$$a = \frac{3}{2}$$ Alternatively, we know that for an exponential function, the ratio of consecutive terms $f(x+1)/f(x)$ is constant and equal to $a$. So, $a = f(3)/f(2) = 18/12 = 3/2$. We want to find $f(1)$. We know that $f(2)/f(1) = a$. Therefore, $f(1) = f(2)/a$. $$f(1) = \frac{12}{3/2} = 12 \cdot \frac{2}{3} = 8$$ So the correct answer is 8.
18. Solve the equation $3^{x+1} = 27^{x-1}$.
    Answer: 2 To solve this exponential equation, we need to express both sides with the same base. We can write 27 as a power of 3, since $27 = 3^3$. Substitute this into the equation: $3^{x+1} = (3^3)^{x-1}$ Using the exponent rule $(a^m)^n = a^{mn}$, we get: $3^{x+1} = 3^{3(x-1)}$ $3^{x+1} = 3^{3x-3}$ Now that the bases are equal, we can equate the exponents: $x+1 = 3x-3$ Now, we solve this linear equation for $x$: $1+3 = 3x-x$ $4 = 2x$ $x=2$.

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