IB Mathematics: Applications and Interpretation Quiz: Logarithmic Scales And Transformations
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Logarithmic Scales And TransformationsQuestion 1 of 20

A set of bivariate data (x,y)(x, y) is believed to follow a power law relationship of the form y=axby=ax^b. Which of the following transformations would result in a linear graph, and what would the gradient of that graph represent?

Plotting yy against log(x)\log(x) gives a linear graph with gradient bb.
Plotting log(y)\log(y) against xx gives a linear graph with gradient log(b)\log(b).
Plotting log(y)\log(y) against log(x)\log(x) gives a linear graph with gradient aa.
Plotting log(y)\log(y) against log(x)\log(x) gives a linear graph with gradient bb.
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Logarithmic Scales And Transformations

Practice Logarithmic Scales And Transformations in IB Mathematics: Applications and Interpretation 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 Logarithmic Scales And Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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.

All questions

Question 1

A set of bivariate data (x,y)(x, y) is believed to follow a power law relationship of the form y=axby=ax^b. Which of the following transformations would result in a linear graph, and what would the gradient of that graph represent?

  1. Plotting yy against log(x)\log(x) gives a linear graph with gradient bb.
  2. Plotting log(y)\log(y) against xx gives a linear graph with gradient log(b)\log(b).
  3. Plotting log(y)\log(y) against log(x)\log(x) gives a linear graph with gradient aa.
  4. Plotting log(y)\log(y) against log(x)\log(x) gives a linear graph with gradient bb. (correct answer)
Explanation: For a power law relationship y=axby=ax^b, we take the logarithm of both sides to linearize it. Using a logarithm of any base (e.g., base 10 or base e), we get log(y)=log(axb)=log(a)+log(xb)=log(a)+blog(x)\log(y) = \log(ax^b) = \log(a) + \log(x^b) = \log(a) + b\log(x). This equation is in the form of a straight line, Y=c+mXY = c + mX, where Y=log(y)Y = \log(y), X=log(x)X = \log(x), the y-intercept is c=log(a)c = \log(a), and the gradient is m=bm = b. Therefore, plotting log(y)\log(y) against log(x)\log(x) results in a linear graph with a gradient of bb.

Question 2

A population PP is modelled by the function P(t)=A(100.5)tP(t) = A \cdot (10^{0.5})^t, where tt is time in years. The data is linearized by plotting log10(P)\log_{10}(P) versus tt. What is the gradient of the resulting line?

  1. 0.5 (correct answer)
  2. 1.15
  3. 3.16
  4. 10
Explanation: The model is P(t)=A(100.5)tP(t) = A \cdot (10^{0.5})^t. This can be rewritten using exponent rules as P(t)=A100.5tP(t) = A \cdot 10^{0.5t}. To linearize this by plotting log10(P)\log_{10}(P) versus tt, we take the logarithm base 10 of both sides: log10(P)=log10(A100.5t)\log_{10}(P) = \log_{10}(A \cdot 10^{0.5t}). Using the product rule for logarithms: log10(P)=log10(A)+log10(100.5t)\log_{10}(P) = \log_{10}(A) + \log_{10}(10^{0.5t}). Using the property logb(bx)=x\log_b(b^x) = x, this simplifies to log10(P)=log10(A)+0.5t\log_{10}(P) = \log_{10}(A) + 0.5t. This is a linear equation of the form Y=c+mtY = c + mt, where Y=log10(P)Y = \log_{10}(P), tt is the x-variable, the y-intercept is c=log10(A)c = \log_{10}(A), and the gradient is m=0.5m = 0.5.

Question 3

A student is modelling a relationship with the function y=4(1.5)xy = 4 \cdot (1.5)^x. To linearize the data, they decide to plot ln(y)\ln(y) against xx. What is the theoretical gradient of this linear graph, correct to 3 significant figures?

  1. 0.176
  2. 0.405 (correct answer)
  3. 1.39
  4. 1.50
Explanation: The function is an exponential model of the form y=abxy=ab^x, with a=4a=4 and b=1.5b=1.5. To linearize this model, we take the natural logarithm of both sides: ln(y)=ln(4(1.5)x)\ln(y) = \ln(4 \cdot (1.5)^x). Using the product rule for logarithms, this becomes ln(y)=ln(4)+ln((1.5)x)\ln(y) = \ln(4) + \ln((1.5)^x). Using the power rule for logarithms, we get ln(y)=ln(4)+xln(1.5)\ln(y) = \ln(4) + x\ln(1.5). This equation is in the form of a straight line, Y=c+mxY = c + mx, where Y=ln(y)Y=\ln(y). The y-intercept is c=ln(4)c = \ln(4) and the gradient is m=ln(1.5)m = \ln(1.5). Calculating the gradient: m=ln(1.5)0.405465...m = \ln(1.5) \approx 0.405465.... Correct to 3 significant figures, the gradient is 0.405.

Question 4

The growth of a bacterial colony is modelled by the exponential function P=P0ektP = P_0 e^{kt}, where PP is the population at time tt hours. To analyse the growth rate, a graph of ln(P)\ln(P) is plotted against tt. The regression line has a gradient of 0.05. What is the correct interpretation of this gradient?

  1. The initial population of the colony was 0.05.
  2. The population increases by a factor of 0.05 each hour.
  3. The population grows at a continuous rate of 5% per hour. (correct answer)
  4. The population's doubling time is 0.05 hours.
Explanation: The model is P=P0ektP = P_0 e^{kt}. To linearize this, we take the natural logarithm of both sides: ln(P)=ln(P0ekt)=ln(P0)+ln(ekt)=ln(P0)+kt\ln(P) = \ln(P_0 e^{kt}) = \ln(P_0) + \ln(e^{kt}) = \ln(P_0) + kt. This is a linear equation in the form Y=c+mtY = c + mt, where Y=ln(P)Y = \ln(P), the y-intercept c=ln(P0)c = \ln(P_0), and the gradient m=km = k. The given gradient is 0.05, so k=0.05k=0.05. In the model P=P0ektP = P_0 e^{kt}, the parameter kk represents the continuous relative growth rate. A value of k=0.05k=0.05 corresponds to a continuous growth rate of 5% per hour.

Question 5

A researcher is studying the energy released by earthquakes, which can range from very small tremors to catastrophic events. Which of the following is the primary mathematical reason for using a logarithmic scale, like the Richter scale, to represent this data?

  1. To convert a non-linear relationship into a linear one for easier regression analysis.
  2. To effectively represent and compare values that span many orders of magnitude on a single, comprehensible scale. (correct answer)
  3. To ensure that the mean and median of the energy data are closer together, making the distribution more symmetric.
  4. To force the correlation coefficient of the data to be closer to 1 or -1.
Explanation: The energy released by earthquakes varies immensely, over many powers of 10 (orders of magnitude). A linear scale would be impractical, as small but significant tremors would be indistinguishable from zero if the scale also had to accommodate the largest earthquakes. A logarithmic scale compresses this vast range into a more manageable set of numbers (e.g., 1 to 10). This makes it possible to represent and compare all values on a single, meaningful scale. While log transformations are used for linearization (A) and can make distributions more symmetric (C), the fundamental purpose of a scale like Richter's is to handle the enormous range of the underlying data.

Question 6

The relationship between two variables xx and yy is given by the power model y=axky=ax^k. When log10(y)\log_{10}(y) is plotted against log10(x)\log_{10}(x), a straight line is formed which passes through the points (0.5,2.1)(0.5, 2.1) and (1.5,5.1)(1.5, 5.1). Find the value of the exponent kk.

  1. 0.33
  2. 3.00 (correct answer)
  3. 3.98
  4. 20.09
Explanation: The power model is y=axky=ax^k. Taking the logarithm base 10 of both sides gives log10(y)=log10(a)+klog10(x)\log_{10}(y) = \log_{10}(a) + k\log_{10}(x). This is a linear equation of the form Y=c+mXY = c + mX, where Y=log10(y)Y = \log_{10}(y), X=log10(x)X = \log_{10}(x), the y-intercept c=log10(a)c = \log_{10}(a), and the gradient m=km = k. We are given two points on this line: (X1,Y1)=(0.5,2.1)(X_1, Y_1)=(0.5, 2.1) and (X2,Y2)=(1.5,5.1)(X_2, Y_2)=(1.5, 5.1). The gradient of this line is k=m=Y2Y1X2X1=5.12.11.50.5=3.01.0=3k = m = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{5.1 - 2.1}{1.5 - 0.5} = \frac{3.0}{1.0} = 3. Thus, the value of the exponent kk is 3.00.

Question 7

The relationship between the mass MM (in kg) and the wingspan WW (in m) of a species of bird is modelled by the equation M=aWbM = aW^b. A researcher plots log10(M)\log_{10}(M) against log10(W)\log_{10}(W) and obtains a line of best fit with the equation Y=2.5X0.7Y = 2.5X - 0.7, where Y=log10(M)Y = \log_{10}(M) and X=log10(W)X = \log_{10}(W). Find the value of aa, correct to two decimal places.

  1. -0.70
  2. 0.20 (correct answer)
  3. 0.50
  4. 2.50
Explanation: The original model is the power law M=aWbM = aW^b. Taking the common logarithm (base 10) of both sides gives log10(M)=log10(aWb)=log10(a)+log10(Wb)=log10(a)+blog10(W)\log_{10}(M) = \log_{10}(aW^b) = \log_{10}(a) + \log_{10}(W^b) = \log_{10}(a) + b\log_{10}(W). This equation has the linear form Y=c+mXY = c + mX, where Y=log10(M)Y = \log_{10}(M), X=log10(W)X = \log_{10}(W), the gradient m=bm = b, and the y-intercept c=log10(a)c = \log_{10}(a). The given regression line is Y=2.5X0.7Y = 2.5X - 0.7. Comparing these, we find that b=2.5b=2.5 and the y-intercept log10(a)=0.7\log_{10}(a) = -0.7. To find aa, we calculate a=100.70.1995...a = 10^{-0.7} \approx 0.1995.... Rounded to two decimal places, a=0.20a = 0.20.

Question 8

A chemist has a solution with a pH of 5. She adds a substance that causes the hydrogen ion concentration, [H+][H^+], to decrease by a factor of 1000. What is the new pH of the solution?

  1. 2
  2. 5.003
  3. 8 (correct answer)
  4. 15
Explanation: The pH of a solution is defined as pH=log10[H+]pH = -\log_{10}[H^+]. Let the initial pH be pH1=5pH_1 = 5 and the initial concentration be [H1+][H^+_1]. Let the final pH be pH2pH_2 and the final concentration be [H2+][H^+_2]. We have [H2+]=[H1+]/1000[H^+_2] = [H^+_1] / 1000. The change in pH is ΔpH=pH2pH1=log10[H2+](log10[H1+])=log10[H1+]log10[H2+]=log10([H1+]/[H2+])\Delta pH = pH_2 - pH_1 = -\log_{10}[H^+_2] - (-\log_{10}[H^+_1]) = \log_{10}[H^+_1] - \log_{10}[H^+_2] = \log_{10}([H^+_1]/[H^+_2]). Since [H1+]/[H2+]=1000[H^+_1]/[H^+_2] = 1000, the change in pH is log10(1000)=3\log_{10}(1000) = 3. So, pH2pH1=3pH_2 - pH_1 = 3. Since pH1=5pH_1 = 5, the new pH is pH2=5+3=8pH_2 = 5 + 3 = 8. A decrease in [H+][H^+] (less acidic) corresponds to an increase in pH.

Question 9

An earthquake in region A was measured to have a magnitude of 6.8 on the Richter scale. A seismic event in region B was found to be 40 times less intense than the earthquake in region A. What is the magnitude of the seismic event in region B on the Richter scale, correct to one decimal place?

  1. 0.2
  2. 5.2 (correct answer)
  3. 6.6
  4. 8.4
Explanation: The Richter scale magnitude MM is defined as M=log10(I/S)M = \log_{10}(I/S), where II is the intensity of the earthquake and SS is a standard intensity. For region A, MA=6.8=log10(IA/S)M_A = 6.8 = \log_{10}(I_A/S). For region B, the intensity is IB=IA/40I_B = I_A / 40. The magnitude in region B is MB=log10(IB/S)=log10((IA/40)/S)M_B = \log_{10}(I_B/S) = \log_{10}((I_A/40)/S). Using the logarithm property log(x/y)=log(x)log(y)\log(x/y) = \log(x) - \log(y), we can write MB=log10(IA/S)log10(40)M_B = \log_{10}(I_A/S) - \log_{10}(40). We know log10(IA/S)=6.8\log_{10}(I_A/S) = 6.8, so MB=6.8log10(40)6.81.602=5.198M_B = 6.8 - \log_{10}(40) \approx 6.8 - 1.602 = 5.198. Rounded to one decimal place, the magnitude is 5.2.

Question 10

A biologist models the growth of a yeast culture using the equation P(t)=ABtP(t) = A \cdot B^t, where PP is the population at time tt hours. To linearize the data, the biologist plots ln(P)\ln(P) against tt. The resulting line of best fit is given by the equation Y=0.15t+4.20Y = 0.15t + 4.20, where Y=ln(P)Y = \ln(P). What is the value of AA in the biologist's model, correct to the nearest integer?

  1. 1.16
  2. 4.20
  3. 67 (correct answer)
  4. 15849
Explanation: The original model is P(t)=ABtP(t) = A \cdot B^t. Taking the natural logarithm of both sides gives ln(P)=ln(ABt)=ln(A)+ln(Bt)=ln(A)+tln(B)\ln(P) = \ln(A \cdot B^t) = \ln(A) + \ln(B^t) = \ln(A) + t \ln(B). This equation is in the form Y=mt+cY = mt + c, where Y=ln(P)Y = \ln(P), the gradient m=ln(B)m = \ln(B), and the y-intercept c=ln(A)c = \ln(A). The given line of best fit is Y=0.15t+4.20Y = 0.15t + 4.20. By comparing the two forms, we can see that the y-intercept ln(A)=4.20\ln(A) = 4.20. To find AA, we solve for A=e4.2066.686A = e^{4.20} \approx 66.686. Rounded to the nearest integer, A=67A = 67.

Question 11

An earthquake in region A was measured to have a magnitude of 6.8 on the Richter scale. A seismic event in region B was found to be 40 times less intense than the earthquake in region A. What is the magnitude of the seismic event in region B on the Richter scale, correct to one decimal place?

  1. 0.2
  2. 5.2 (correct answer)
  3. 6.6
  4. 8.4
Explanation: The Richter scale magnitude MM is defined as M=log10(I/S)M = \log_{10}(I/S), where II is the intensity of the earthquake and SS is a standard intensity. For region A, MA=6.8=log10(IA/S)M_A = 6.8 = \log_{10}(I_A/S). For region B, the intensity is IB=IA/40I_B = I_A / 40. The magnitude in region B is MB=log10(IB/S)=log10((IA/40)/S)M_B = \log_{10}(I_B/S) = \log_{10}((I_A/40)/S). Using the logarithm property log(x/y)=log(x)log(y)\log(x/y) = \log(x) - \log(y), we can write MB=log10(IA/S)log10(40)M_B = \log_{10}(I_A/S) - \log_{10}(40). We know log10(IA/S)=6.8\log_{10}(I_A/S) = 6.8, so MB=6.8log10(40)6.81.602=5.198M_B = 6.8 - \log_{10}(40) \approx 6.8 - 1.602 = 5.198. Rounded to one decimal place, the magnitude is 5.2.

Question 12

A dataset is modelled by the exponential function f(x)=abxf(x) = ab^x. The data is linearized by plotting ln(y)\ln(y) against xx. The resulting line passes through the points (2,5.5)(2, 5.5) and (6,8.3)(6, 8.3). Find the value of bb in the model, correct to two decimal places.

  1. 0.70
  2. 2.01 (correct answer)
  3. 5.01
  4. 60.34
Explanation: The exponential model is y=abxy = ab^x. Linearizing by taking the natural logarithm gives ln(y)=ln(a)+xln(b)\ln(y) = \ln(a) + x\ln(b). This is a linear equation of the form Y=c+mxY = c + mx, where Y=ln(y)Y = \ln(y), the y-intercept c=ln(a)c = \ln(a), and the gradient m=ln(b)m = \ln(b). The gradient of the line passing through (x1,Y1)=(2,5.5)(x_1, Y_1) = (2, 5.5) and (x2,Y2)=(6,8.3)(x_2, Y_2) = (6, 8.3) is m=Y2Y1x2x1=8.35.562=2.84=0.7m = \frac{Y_2 - Y_1}{x_2 - x_1} = \frac{8.3 - 5.5}{6 - 2} = \frac{2.8}{4} = 0.7. Since the gradient m=ln(b)m = \ln(b), we have ln(b)=0.7\ln(b) = 0.7. To find bb, we calculate b=e0.72.01375...b = e^{0.7} \approx 2.01375.... Correct to two decimal places, b=2.01b = 2.01.

Question 13

A scientist proposes a power law model S=aTbS = aT^b to relate the number of species SS on an island to its area TT in square kilometers. By plotting ln(S)\ln(S) versus ln(T)\ln(T), she obtains the regression line Y=0.25X+1.8Y = 0.25X + 1.8, where Y=ln(S)Y=\ln(S) and X=ln(T)X=\ln(T). Predict the number of species on an island with an area of 200 square kilometers, rounding your answer to the nearest whole number.

  1. 3
  2. 23 (correct answer)
  3. 52
  4. 237
Explanation: The linearized equation is ln(S)=0.25ln(T)+1.8\ln(S) = 0.25\ln(T) + 1.8. We want to predict SS for T=200T = 200. First, we calculate X=ln(T)=ln(200)5.2983X = \ln(T) = \ln(200) \approx 5.2983. Next, we substitute this into the regression equation to find Y=ln(S)Y = \ln(S): ln(S)=0.25×5.2983+1.81.3246+1.8=3.1246\ln(S) = 0.25 \times 5.2983 + 1.8 \approx 1.3246 + 1.8 = 3.1246. To find SS, we take the exponent of this value: S=e3.124622.748S = e^{3.1246} \approx 22.748. Rounded to the nearest whole number, the predicted number of species is 23.

Question 14

A power law model y=axby = ax^b is linearized by plotting YY against XX, where Y=ln(y)Y=\ln(y) and X=ln(x)X=\ln(x). The equation of the resulting regression line is Y=mX+cY = mX + c. Which expression correctly gives the value of the coefficient aa from the original model?

  1. a=ca=c
  2. a=eca=e^c (correct answer)
  3. a=ma=m
  4. a=ema=e^m
Explanation: The power law model is y=axby = ax^b. To linearize it, we take the natural logarithm of both sides: ln(y)=ln(axb)\ln(y) = \ln(ax^b). Using logarithm properties, this becomes ln(y)=ln(a)+ln(xb)\ln(y) = \ln(a) + \ln(x^b), which simplifies to ln(y)=ln(a)+bln(x)\ln(y) = \ln(a) + b\ln(x). This equation has the form of a straight line, Y=c+mXY = c + mX, where Y=ln(y)Y=\ln(y), X=ln(x)X=\ln(x). By comparing the terms, we see that the y-intercept cc corresponds to ln(a)\ln(a) and the gradient mm corresponds to bb. To find aa from c=ln(a)c = \ln(a), we need to apply the inverse operation of the natural logarithm, which is exponentiation with base ee. Therefore, a=eca = e^c.

Question 15

The growth of a bacterial colony is modelled by the exponential function P=P0ektP = P_0 e^{kt}, where PP is the population at time tt hours. To analyse the growth rate, a graph of ln(P)\ln(P) is plotted against tt. The regression line has a gradient of 0.05. What is the correct interpretation of this gradient?

  1. The initial population of the colony was 0.05.
  2. The population increases by a factor of 0.05 each hour.
  3. The population grows at a continuous rate of 5% per hour. (correct answer)
  4. The population's doubling time is 0.05 hours.
Explanation: The model is P=P0ektP = P_0 e^{kt}. To linearize this, we take the natural logarithm of both sides: ln(P)=ln(P0ekt)=ln(P0)+ln(ekt)=ln(P0)+kt\ln(P) = \ln(P_0 e^{kt}) = \ln(P_0) + \ln(e^{kt}) = \ln(P_0) + kt. This is a linear equation in the form Y=c+mtY = c + mt, where Y=ln(P)Y = \ln(P), the y-intercept c=ln(P0)c = \ln(P_0), and the gradient m=km = k. The given gradient is 0.05, so k=0.05k=0.05. In the model P=P0ektP = P_0 e^{kt}, the parameter kk represents the continuous relative growth rate. A value of k=0.05k=0.05 corresponds to a continuous growth rate of 5% per hour.

Question 16

A scientist collects data on a variable yy that depends on xx. To determine the nature of the relationship, two transformations are performed. A plot of ln(y)\ln(y) versus xx has a Pearson's correlation coefficient of r=0.88r = 0.88. A plot of ln(y)\ln(y) versus ln(x)\ln(x) has a Pearson's correlation coefficient of r=0.99r = -0.99. Which of the following is the most appropriate conclusion?

  1. An exponential model of the form y=abxy=ab^x with b>1b>1 is the best fit for the data.
  2. An exponential model of the form y=abxy=ab^x with 0<b<10<b<1 is the best fit for the data.
  3. A power law model of the form y=axby=ax^b with b>0b>0 is the best fit for the data.
  4. A power law model of the form y=axby=ax^b with b<0b<0 is the best fit for the data. (correct answer)
Explanation: The Pearson's correlation coefficient, rr, measures the strength and direction of a linear relationship. A value closer to 11 or 1-1 indicates a stronger linear fit. Here, r=0.99r = -0.99 is much closer to 1-1 than r=0.88r = 0.88 is to 11. This means the second transformation provides a better linear model for the data. The transformation of plotting ln(y)\ln(y) versus ln(x)\ln(x) linearizes a power law relationship of the form y=axby=ax^b. The gradient of the line ln(y)=bln(x)+ln(a)\ln(y) = b\ln(x) + \ln(a) is bb. Since the correlation coefficient rr is negative, the slope of the regression line must be negative, which implies that the exponent bb is negative. Thus, a power law model with b<0b<0 is the best fit.

Question 17

A student is modelling a relationship with the function y=4(1.5)xy = 4 \cdot (1.5)^x. To linearize the data, they decide to plot ln(y)\ln(y) against xx. What is the theoretical gradient of this linear graph, correct to 3 significant figures?

  1. 0.176
  2. 0.405 (correct answer)
  3. 1.39
  4. 1.50
Explanation: The function is an exponential model of the form y=abxy=ab^x, with a=4a=4 and b=1.5b=1.5. To linearize this model, we take the natural logarithm of both sides: ln(y)=ln(4(1.5)x)\ln(y) = \ln(4 \cdot (1.5)^x). Using the product rule for logarithms, this becomes ln(y)=ln(4)+ln((1.5)x)\ln(y) = \ln(4) + \ln((1.5)^x). Using the power rule for logarithms, we get ln(y)=ln(4)+xln(1.5)\ln(y) = \ln(4) + x\ln(1.5). This equation is in the form of a straight line, Y=c+mxY = c + mx, where Y=ln(y)Y=\ln(y). The y-intercept is c=ln(4)c = \ln(4) and the gradient is m=ln(1.5)m = \ln(1.5). Calculating the gradient: m=ln(1.5)0.405465...m = \ln(1.5) \approx 0.405465.... Correct to 3 significant figures, the gradient is 0.405.

Question 18

It is observed for a certain physical phenomenon that when the logarithm of a quantity PP is plotted against the logarithm of a quantity QQ, the result is a straight line with a negative slope. Which statement best describes the relationship between PP and QQ?

  1. As QQ increases by a fixed amount, PP decreases by a fixed amount.
  2. As QQ increases by a fixed amount, PP decreases by a constant percentage.
  3. As QQ increases by a constant percentage, PP decreases by a constant percentage. (correct answer)
  4. As QQ increases by a constant percentage, PP decreases by a fixed amount.
Explanation: A linear relationship between log(P)\log(P) and log(Q)\log(Q) implies a power law relationship between PP and QQ, of the form P=aQbP = aQ^b. A negative slope means that the exponent bb is negative. In a power law relationship, a constant percentage (or multiplicative) change in one variable results in a constant percentage (multiplicative) change in the other. For example, if QQ is multiplied by a factor kk, then PP is multiplied by a factor kbk^b. Since bb is negative, if k>1k>1 (an increase), then 0<kb<10 < k^b < 1 (a decrease). This means that as QQ increases by a constant percentage, PP decreases by a constant percentage. Option A describes a linear relationship. Option B describes an exponential relationship.

Question 19

A biologist models the growth of a yeast culture using the equation P(t)=ABtP(t) = A \cdot B^t, where PP is the population at time tt hours. To linearize the data, the biologist plots ln(P)\ln(P) against tt. The resulting line of best fit is given by the equation Y=0.15t+4.20Y = 0.15t + 4.20, where Y=ln(P)Y = \ln(P). What is the value of AA in the biologist's model, correct to the nearest integer?

  1. 1.16
  2. 4.20
  3. 67 (correct answer)
  4. 15849
Explanation: The original model is P(t)=ABtP(t) = A \cdot B^t. Taking the natural logarithm of both sides gives ln(P)=ln(ABt)=ln(A)+ln(Bt)=ln(A)+tln(B)\ln(P) = \ln(A \cdot B^t) = \ln(A) + \ln(B^t) = \ln(A) + t \ln(B). This equation is in the form Y=mt+cY = mt + c, where Y=ln(P)Y = \ln(P), the gradient m=ln(B)m = \ln(B), and the y-intercept c=ln(A)c = \ln(A). The given line of best fit is Y=0.15t+4.20Y = 0.15t + 4.20. By comparing the two forms, we can see that the y-intercept ln(A)=4.20\ln(A) = 4.20. To find AA, we solve for A=e4.2066.686A = e^{4.20} \approx 66.686. Rounded to the nearest integer, A=67A = 67.

Question 20

The sound level of a machine in a factory is 90 dB. A second, identical machine is turned on next to the first one, which doubles the total sound intensity. What is the new sound level in decibels, correct to one decimal place?

  1. 92.0 dB
  2. 93.0 dB (correct answer)
  3. 100.0 dB
  4. 180.0 dB
Explanation: The sound level LL in decibels (dB) is given by L=10log10(I/I0)L = 10 \log_{10}(I/I_0), where II is the sound intensity and I0I_0 is a reference intensity. Let I1I_1 be the intensity of one machine. We are given that its sound level is 90=10log10(I1/I0)90 = 10 \log_{10}(I_1/I_0). When the second identical machine is turned on, the new intensity is I2=2I1I_2 = 2I_1. The new sound level L2L_2 is L2=10log10(2I1/I0)L_2 = 10 \log_{10}(2I_1/I_0). Using the logarithm property log(ab)=log(a)+log(b)\log(ab) = \log(a) + \log(b), we have L2=10(log10(2)+log10(I1/I0))=10log10(2)+10log10(I1/I0)L_2 = 10 (\log_{10}(2) + \log_{10}(I_1/I_0)) = 10\log_{10}(2) + 10\log_{10}(I_1/I_0). Since 10log10(I1/I0)=9010\log_{10}(I_1/I_0) = 90, the equation becomes L2=10log10(2)+90L_2 = 10\log_{10}(2) + 90. Calculating this gives L210(0.30103)+90=3.0103+90=93.0103L_2 \approx 10(0.30103) + 90 = 3.0103 + 90 = 93.0103. To one decimal place, the new sound level is 93.0 dB.