IB Mathematics: Analysis and Approaches Quiz: Exponential Functions
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Exponential FunctionsQuestion 1 of 20

Find the range of the function f(x)=7−3⋅2x+1f(x) = 7 - 3 \cdot 2^{x+1}.

y>7y > 7
y<−3y < -3
y>−3y > -3
y<7y < 7
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Exponential Functions

Practice Exponential Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exponential Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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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.

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Question 1

Find the range of the function f(x)=7−3⋅2x+1f(x) = 7 - 3 \cdot 2^{x+1}.

  1. y>7y > 7
  2. y<−3y < -3
  3. y>−3y > -3
  4. y<7y < 7 (correct answer)
Explanation: 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⋅2x+1f(x) = 7 - 3 \cdot 2^{x+1} to better see its structure. Since 2x+1=2⋅2x2^{x+1} = 2 \cdot 2^x, we have f(x)=7−6⋅2xf(x) = 7 - 6 \cdot 2^x. Now consider what happens to 2x2^x as xx changes: as x→∞x \to \infty, we get 2x→∞2^x \to \infty, and as x→−∞x \to -\infty, we get 2x→0+2^x \to 0^+ (approaches zero from the positive side). This means 6⋅2x6 \cdot 2^x ranges from values approaching 0 to +∞+\infty. Therefore, f(x)=7−6⋅2xf(x) = 7 - 6 \cdot 2^x ranges from 7−∞=−∞7 - \infty = -\infty up to (but never reaching) 7−0=77 - 0 = 7. The range is y<7y < 7. Looking at the wrong answers: Choice A suggests y>7y > 7, which incorrectly assumes the function increases without bound—this ignores the negative coefficient. Choice B claims y<−3y < -3, likely from incorrectly thinking there's a lower bound at y=−3y = -3. Choice C suggests y>−3y > -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⋅cxa + b \cdot c^x, identify the horizontal asymptote (y=ay = a) and determine whether the function approaches it from above or below based on the sign of bb. When b<0b < 0, the range excludes values above the asymptote.

Question 2

The temperature TT, in degrees Celsius, of a cooling object tt minutes after it is removed from a heat source is given by T(t)=20+64(12)t/5T(t) = 20 + 64{\left(\frac{1}{2}\right)}^{t/5}. What is the temperature of the object after 10 minutes?

  1. 20°C
  2. 36°C (correct answer)
  3. 52°C
  4. 84°C
Explanation: We need to find the value of T(10)T(10). Substitute t=10t=10 into the given function:\nT(10)=20+64(12)10/5T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{10/5}\nFirst, simplify the exponent:\n105=2\frac{10}{5} = 2\nNow the equation becomes:\nT(10)=20+64(12)2T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{2}\nCalculate the power:\n(12)2=14{\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}\nNow substitute this back:\nT(10)=20+64(14)T(10) = 20 + 64\left(\frac{1}{4}\right)\nT(10)=20+16T(10) = 20 + 16\nT(10)=36T(10) = 36\nThe temperature after 10 minutes is 36°C.

Question 3

The temperature TT, in degrees Celsius, of a cooling object tt minutes after it is removed from a heat source is given by T(t)=20+64(12)t/5T(t) = 20 + 64{\left(\frac{1}{2}\right)}^{t/5}. What is the temperature of the object after 10 minutes?

  1. 20°C
  2. 36°C (correct answer)
  3. 52°C
  4. 84°C
Explanation: We need to find the value of T(10)T(10). Substitute t=10t=10 into the given function:\nT(10)=20+64(12)10/5T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{10/5}\nFirst, simplify the exponent:\n105=2\frac{10}{5} = 2\nNow the equation becomes:\nT(10)=20+64(12)2T(10) = 20 + 64{\left(\frac{1}{2}\right)}^{2}\nCalculate the power:\n(12)2=14{\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}\nNow substitute this back:\nT(10)=20+64(14)T(10) = 20 + 64\left(\frac{1}{4}\right)\nT(10)=20+16T(10) = 20 + 16\nT(10)=36T(10) = 36\nThe temperature after 10 minutes is 36°C.

Question 4

The value of a car, VV, in dollars, depreciates according to the model V(t)=30000(0.8)tV(t) = 30000(0.8)^t, where tt 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)?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 7
Explanation: We need to find the smallest integer tt for which V(t)<10000V(t) < 10000.\nSet up the inequality:\n30000(0.8)t<1000030000(0.8)^t < 10000\nDivide both sides by 30000:\n(0.8)t<1000030000(0.8)^t < \frac{10000}{30000}\n(0.8)t<13(0.8)^t < \frac{1}{3}\nThis could be solved with logarithms, but for a multiple choice question, we can test the integer values of tt.\nt=1:(0.8)1=0.8t=1: (0.8)^1 = 0.8. Is 0.8<1/30.8 < 1/3? No, 0.8>0.333...0.8 > 0.333....\nt=2:(0.8)2=0.64t=2: (0.8)^2 = 0.64. Still greater than 1/31/3.\nt=3:(0.8)3=0.512t=3: (0.8)^3 = 0.512. Still greater than 1/31/3.\nt=4:(0.8)4=0.4096t=4: (0.8)^4 = 0.4096. Still greater than 1/31/3.\nt=5:(0.8)5=0.32768t=5: (0.8)^5 = 0.32768. This is less than 1/3≈0.333...1/3 \approx 0.333....\nSo, the first full year when the value is less than ($10000) is t=5t=5.

Question 5

An exponential function of the form f(x)=kaxf(x) = ka^x passes through the points (2,12)(2, 12) and (3,18)(3, 18). Find the value of f(1)f(1).

  1. 6
  2. 163\frac{16}{3}
  3. 8 (correct answer)
  4. 9
Explanation: To find the parameters of f(x)=kaxf(x) = ka^x, we set up a system of two equations using the given points:\n1. ka2=12ka^2 = 12\n2. ka3=18ka^3 = 18\n\nDivide the second equation by the first to eliminate kk:\nka3ka2=1812\frac{ka^3}{ka^2} = \frac{18}{12} a=32a = \frac{3}{2}\nAlternatively, we know that for an exponential function, the ratio of consecutive terms f(x+1)/f(x)f(x+1)/f(x) is constant and equal to aa. So, a=f(3)/f(2)=18/12=3/2a = f(3)/f(2) = 18/12 = 3/2.\n\nWe want to find f(1)f(1). We know that f(2)/f(1)=af(2)/f(1) = a. Therefore, f(1)=f(2)/af(1) = f(2)/a.\nf(1)=123/2=12⋅23=8f(1) = \frac{12}{3/2} = 12 \cdot \frac{2}{3} = 8\nSo the correct answer is 8.

Question 6

A population of bacteria is modelled by P(t)=100atP(t) = 100a^t, where tt is the time in hours. After 2 hours, the population is 900. At what time tt will the population reach 8100?

  1. 3 hours
  2. 4 hours (correct answer)
  3. 6 hours
  4. 20 hours
Explanation: First, we use the given information to find the growth factor aa.\nWe know P(2)=900P(2) = 900. Substituting into the model:\n100a2=900100a^2 = 900\na2=9a^2 = 9\nSince aa must be positive for population growth, a=3a = 3.\nSo the model is P(t)=100(3)tP(t) = 100(3)^t.\n\nNext, we need to find the time tt when the population reaches 8100.\n100(3)t=8100100(3)^t = 8100\n3t=813^t = 81\nWe can write 81 as a power of 3: 81=92=(32)2=3481 = 9^2 = (3^2)^2 = 3^4.\nSo, 3t=343^t = 3^4.\nTherefore, t=4t = 4 hours.

Question 7

Consider the function f(x)=k⋅2−x/3f(x) = k \cdot 2^{-x/3}. The graph of ff passes through the point (6,5)(6, 5). Find the value of f(−3)f(-3).

  1. 10
  2. 20
  3. 30
  4. 40 (correct answer)
Explanation: First, we must find the value of the constant kk using the fact that the graph passes through (6,5)(6, 5). This means f(6)=5f(6)=5.\nSubstitute x=6x=6 and f(x)=5f(x)=5 into the function's equation:\n5=k⋅2−6/35 = k \cdot 2^{-6/3}\n5=k⋅2−25 = k \cdot 2^{-2}\n5=k⋅1225 = k \cdot \frac{1}{2^2}\n5=k⋅145 = k \cdot \frac{1}{4}\nk=5×4=20k = 5 \times 4 = 20.\nNow that we have kk, the function is f(x)=20⋅2−x/3f(x) = 20 \cdot 2^{-x/3}.\nNext, we need to find the value of f(−3)f(-3). Substitute x=−3x=-3 into the function:\nf(−3)=20⋅2−(−3)/3f(-3) = 20 \cdot 2^{-(-3)/3}\nf(−3)=20⋅23/3f(-3) = 20 \cdot 2^{3/3}\nf(−3)=20⋅21f(-3) = 20 \cdot 2^{1}\nf(−3)=20⋅2=40f(-3) = 20 \cdot 2 = 40.

Question 8

The value of a car, VV, in dollars, depreciates according to the model V(t)=30000(0.8)tV(t) = 30000(0.8)^t, where tt 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)?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 7
Explanation: We need to find the smallest integer tt for which V(t)<10000V(t) < 10000.\nSet up the inequality:\n30000(0.8)t<1000030000(0.8)^t < 10000\nDivide both sides by 30000:\n(0.8)t<1000030000(0.8)^t < \frac{10000}{30000}\n(0.8)t<13(0.8)^t < \frac{1}{3}\nThis could be solved with logarithms, but for a multiple choice question, we can test the integer values of tt.\nt=1:(0.8)1=0.8t=1: (0.8)^1 = 0.8. Is 0.8<1/30.8 < 1/3? No, 0.8>0.333...0.8 > 0.333....\nt=2:(0.8)2=0.64t=2: (0.8)^2 = 0.64. Still greater than 1/31/3.\nt=3:(0.8)3=0.512t=3: (0.8)^3 = 0.512. Still greater than 1/31/3.\nt=4:(0.8)4=0.4096t=4: (0.8)^4 = 0.4096. Still greater than 1/31/3.\nt=5:(0.8)5=0.32768t=5: (0.8)^5 = 0.32768. This is less than 1/3≈0.333...1/3 \approx 0.333....\nSo, the first full year when the value is less than ($10000) is t=5t=5.

Question 9

Find the inverse function, f−1(x)f^{-1}(x), of f(x)=ex−3+4f(x) = e^{x-3} + 4.

  1. f−1(x)=ln⁡(x−4)+3f^{-1}(x) = \ln(x-4) + 3 (correct answer)
  2. f−1(x)=ln⁡(x−4)−3f^{-1}(x) = \ln(x-4) - 3
  3. f−1(x)=ln⁡(x+4)+3f^{-1}(x) = \ln(x+4) + 3
  4. f−1(x)=ln⁡(x+4)−3f^{-1}(x) = \ln(x+4) - 3
Explanation: To find the inverse function, we start with the equation y=f(x)y = f(x), then swap xx and yy, and finally solve for yy.\n1. Start with y=ex−3+4y = e^{x-3} + 4.\n2. Swap xx and yy: x=ey−3+4x = e^{y-3} + 4.\n3. Solve for yy. First, isolate the exponential term:\nx−4=ey−3x - 4 = e^{y-3}\n4. To bring the exponent down, take the natural logarithm of both sides:\nln⁡(x−4)=ln⁡(ey−3)\ln(x-4) = \ln(e^{y-3})\nln⁡(x−4)=y−3\ln(x-4) = y-3\n5. Isolate yy:\ny=ln⁡(x−4)+3y = \ln(x-4) + 3.\nTherefore, the inverse function is f−1(x)=ln⁡(x−4)+3f^{-1}(x) = \ln(x-4) + 3.

Question 10

A population of bacteria is modelled by P(t)=100atP(t) = 100a^t, where tt is the time in hours. After 2 hours, the population is 900. At what time tt will the population reach 8100?

  1. 3 hours
  2. 4 hours (correct answer)
  3. 6 hours
  4. 20 hours
Explanation: First, we use the given information to find the growth factor aa.\nWe know P(2)=900P(2) = 900. Substituting into the model:\n100a2=900100a^2 = 900\na2=9a^2 = 9\nSince aa must be positive for population growth, a=3a = 3.\nSo the model is P(t)=100(3)tP(t) = 100(3)^t.\n\nNext, we need to find the time tt when the population reaches 8100.\n100(3)t=8100100(3)^t = 8100\n3t=813^t = 81\nWe can write 81 as a power of 3: 81=92=(32)2=3481 = 9^2 = (3^2)^2 = 3^4.\nSo, 3t=343^t = 3^4.\nTherefore, t=4t = 4 hours.

Question 11

Find the smallest integer xx for which 3⋅2x>3003 \cdot 2^x > 300.

  1. 6
  2. 7 (correct answer)
  3. 8
  4. 9
Explanation: First, we simplify the inequality:\n3⋅2x>3003 \cdot 2^x > 300\nDivide both sides by 3:\n2x>1002^x > 100\nWe need to find the smallest integer xx that satisfies this condition. We can test powers of 2:\n21=22^1 = 2\n22=42^2 = 4\n23=82^3 = 8\n24=162^4 = 16\n25=322^5 = 32\n26=642^6 = 64 (This is less than 100, so x=6x=6 is not the solution).\n27=1282^7 = 128 (This is greater than 100).\nTherefore, the smallest integer value of xx for which 2x>1002^x > 100 is x=7x=7.

Question 12

What is the solution to the equation 4x−6⋅2x−16=04^{x} - 6 \cdot 2^{x} - 16 = 0?

  1. x=−1x = -1
  2. x=1x = 1
  3. x=3x = 3 (correct answer)
  4. x=8x = 8
Explanation: This equation is a quadratic in disguise. We can rewrite 4x4^x as (22)x=(2x)2(2^2)^x = (2^x)^2. Let y=2xy = 2^x. The equation becomes:\ny2−6y−16=0y^2 - 6y - 16 = 0\nWe can factor this quadratic equation:\n(y−8)(y+2)=0(y-8)(y+2) = 0\nThis gives two possible solutions for yy: y=8y=8 or y=−2y=-2.\n\nNow we substitute back y=2xy = 2^x:\nCase 1: 2x=82^x = 8. Since 8=238 = 2^3, we have x=3x=3.\nCase 2: 2x=−22^x = -2. The exponential function 2x2^x is always positive for real xx, so this case yields no real solution.\n\nThe only solution is x=3x=3.

Question 13

Find the range of the function f(x)=7−3⋅2x+1f(x) = 7 - 3 \cdot 2^{x+1}.

  1. y>7y > 7
  2. y<−3y < -3
  3. y>−3y > -3
  4. y<7y < 7 (correct answer)
Explanation: 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⋅2x+1f(x) = 7 - 3 \cdot 2^{x+1} to better see its structure. Since 2x+1=2⋅2x2^{x+1} = 2 \cdot 2^x, we have f(x)=7−6⋅2xf(x) = 7 - 6 \cdot 2^x. Now consider what happens to 2x2^x as xx changes: as x→∞x \to \infty, we get 2x→∞2^x \to \infty, and as x→−∞x \to -\infty, we get 2x→0+2^x \to 0^+ (approaches zero from the positive side). This means 6⋅2x6 \cdot 2^x ranges from values approaching 0 to +∞+\infty. Therefore, f(x)=7−6⋅2xf(x) = 7 - 6 \cdot 2^x ranges from 7−∞=−∞7 - \infty = -\infty up to (but never reaching) 7−0=77 - 0 = 7. The range is y<7y < 7. Looking at the wrong answers: Choice A suggests y>7y > 7, which incorrectly assumes the function increases without bound—this ignores the negative coefficient. Choice B claims y<−3y < -3, likely from incorrectly thinking there's a lower bound at y=−3y = -3. Choice C suggests y>−3y > -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⋅cxa + b \cdot c^x, identify the horizontal asymptote (y=ay = a) and determine whether the function approaches it from above or below based on the sign of bb. When b<0b < 0, the range excludes values above the asymptote.

Question 14

A substance decays according to the model M(t)=M0(0.85)tM(t) = M_0 (0.85)^t, where tt is in years. What is the annual percentage decay rate of the substance?

  1. 0.15%
  2. 0.85%
  3. 15% (correct answer)
  4. 85%
Explanation: The general model for exponential decay is M(t)=M0(1−r)tM(t) = M_0 (1-r)^t, where rr is the annual decay rate. The term (1−r)(1-r) is the decay factor.\nIn the given model, M(t)=M0(0.85)tM(t) = M_0 (0.85)^t, the decay factor is 0.85.\nSo, we have 1−r=0.851-r = 0.85.\nSolving for rr:\nr=1−0.85=0.15r = 1 - 0.85 = 0.15.\nTo express this rate as a percentage, we multiply by 100:\n0.15×100%=15%0.15 \times 100\% = 15\%.\nThe annual percentage decay rate is 15%.

Question 15

The graph of y=2xy=2^x is reflected in the y-axis and then translated 3 units to the left. What is the equation of the transformed graph?

  1. y=2−x−3y = 2^{-x-3} (correct answer)
  2. y=2−x+3y = 2^{-x+3}
  3. y=−2x−3y = -2^{x-3}
  4. y=2x+3y = 2^{x+3}
Explanation: Let's apply the transformations step-by-step to the function f(x)=2xf(x) = 2^x.\n1. Reflection in the y-axis: This transformation replaces xx with −x-x. The new function is g(x)=f(−x)=2−xg(x) = f(-x) = 2^{-x}.\n2. Translation 3 units to the left: This transformation replaces xx with (x+3)(x+3) in the current function, g(x)g(x). The final function is h(x)=g(x+3)=2−(x+3)h(x) = g(x+3) = 2^{-(x+3)}.\n3. Simplify the exponent: h(x)=2−x−3h(x) = 2^{-x-3}.\nThe equation of the transformed graph is y=2−x−3y = 2^{-x-3}.\nDistractor B arises from applying the transformations in the wrong order: translating left first gives 2x+32^{x+3}, and then reflecting in the y-axis gives 2−x+32^{-x+3}.

Question 16

What is the solution to the equation 4x−6⋅2x−16=04^{x} - 6 \cdot 2^{x} - 16 = 0?

  1. x=−1x = -1
  2. x=1x = 1
  3. x=3x = 3 (correct answer)
  4. x=8x = 8
Explanation: This equation is a quadratic in disguise. We can rewrite 4x4^x as (22)x=(2x)2(2^2)^x = (2^x)^2. Let y=2xy = 2^x. The equation becomes:\ny2−6y−16=0y^2 - 6y - 16 = 0\nWe can factor this quadratic equation:\n(y−8)(y+2)=0(y-8)(y+2) = 0\nThis gives two possible solutions for yy: y=8y=8 or y=−2y=-2.\n\nNow we substitute back y=2xy = 2^x:\nCase 1: 2x=82^x = 8. Since 8=238 = 2^3, we have x=3x=3.\nCase 2: 2x=−22^x = -2. The exponential function 2x2^x is always positive for real xx, so this case yields no real solution.\n\nThe only solution is x=3x=3.

Question 17

The function f(x)=4⋅5x−1+3f(x) = 4 \cdot 5^{x-1} + 3 is obtained from the graph of g(x)=5xg(x) = 5^x by a sequence of transformations. Which of the following describes a correct sequence?

  1. A vertical stretch by a factor of 4, followed by a translation by the vector (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix}. (correct answer)
  2. A translation by the vector (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix}, followed by a vertical stretch by a factor of 4.
  3. A vertical stretch by a factor of 4, followed by a translation by the vector (−13)\begin{pmatrix} -1 \\ 3 \end{pmatrix}.
  4. A translation by the vector (−13)\begin{pmatrix} -1 \\ 3 \end{pmatrix}, followed by a vertical stretch by a factor of 4.
Explanation: The equation is f(x)=4⋅5x−1+3f(x) = 4 \cdot 5^{x-1} + 3. Let's analyze the transformations from g(x)=5xg(x)=5^x.\nThe term x−1x-1 in the exponent corresponds to a horizontal translation of 1 unit to the right. \nThe multiplication by 4 corresponds to a vertical stretch by a factor of 4. \nThe addition of 3 corresponds to a vertical translation of 3 units up.\n\nA horizontal translation of 1 unit right and a vertical translation of 3 units up can be represented by the translation vector (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix}.\n\nThe order of transformations matters. The vertical stretch must be applied to the function 5x−15^{x-1} before the vertical translation is applied.\nLet's check option A: Start with 5x5^x. Apply a vertical stretch by factor 4 to get 4⋅5x4 \cdot 5^x. Then translate by (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix} (1 right, 3 up) to get 4⋅5x−1+34 \cdot 5^{x-1} + 3. This is correct.\nLet's check option B: Start with 5x5^x. Translate by (13)\begin{pmatrix} 1 \\ 3 \end{pmatrix} to get 5x−1+35^{x-1} + 3. Then apply a vertical stretch by factor 4 to get 4(5x−1+3)=4⋅5x−1+124(5^{x-1} + 3) = 4 \cdot 5^{x-1} + 12, which is incorrect.

Question 18

Solve the equation 3x+1=27x−13^{x+1} = 27^{x-1}.

  1. -1
  2. 1
  3. 2 (correct answer)
  4. 4
Explanation: 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=3327 = 3^3.\nSubstitute this into the equation:\n3x+1=(33)x−13^{x+1} = (3^3)^{x-1}\nUsing the exponent rule (am)n=amn(a^m)^n = a^{mn}, we get:\n3x+1=33(x−1)3^{x+1} = 3^{3(x-1)}\n3x+1=33x−33^{x+1} = 3^{3x-3}\nNow that the bases are equal, we can equate the exponents:\nx+1=3x−3x+1 = 3x-3\nNow, we solve this linear equation for xx:\n1+3=3x−x1+3 = 3x-x\n4=2x4 = 2x\nx=2x=2.

Question 19

What is the y-intercept of the graph of f(x)=5(3x+1−2)f(x) = 5(3^{x+1} - 2)?

  1. -10
  2. 3
  3. 5 (correct answer)
  4. 13
Explanation: The y-intercept of a function occurs where the graph crosses the y-axis, which is at x=0x=0. To find the y-intercept, we need to calculate f(0)f(0).\nSubstitute x=0x=0 into the function's equation:\nf(0)=5(30+1−2)f(0) = 5(3^{0+1} - 2)\nf(0)=5(31−2)f(0) = 5(3^1 - 2)\nf(0)=5(3−2)f(0) = 5(3 - 2)\nf(0)=5(1)f(0) = 5(1)\nf(0)=5f(0) = 5.\nThe y-intercept is at (0,5)(0, 5). The value is 5.

Question 20

Given f(x)=32xf(x) = 3^{2x} and g(x)=x+1g(x) = x+1, find the solution to the equation (f∘g)(x)=19(f \circ g)(x) = \frac{1}{9}.

  1. -2 (correct answer)
  2. -3/2
  3. -1
  4. 0
Explanation: First, we find the composite function (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)).\nSubstitute g(x)g(x) into f(x)f(x):\nf(g(x))=32(g(x))=32(x+1)f(g(x)) = 3^{2(g(x))} = 3^{2(x+1)}.\nDistribute the 2 in the exponent: f(g(x))=32x+2f(g(x)) = 3^{2x+2}.\nNow, set this equal to 19\frac{1}{9}:\n32x+2=193^{2x+2} = \frac{1}{9}.\nTo solve this, we need to express both sides with the same base. Since 9=329 = 3^2, we have 19=132=3−2\frac{1}{9} = \frac{1}{3^2} = 3^{-2}.\nSo the equation becomes:\n32x+2=3−23^{2x+2} = 3^{-2}.\nNow we can equate the exponents:\n2x+2=−22x+2 = -2\n2x=−42x = -4\nx=−2x = -2.