Logarithmic Scales & Transformations - IB Mathematics Applications and Interpretation

1. In astronomy, the apparent magnitude, $m$, of a celestial object is related to its brightness, $B$, by $m = -2.5 \log_{10}(B/B_0)$, where $B_0$ is a reference brightness. Star A is 100 times brighter than Star B. What is the difference in their magnitudes, $m_B - m_A$?
   Answer: 5.0 We are given the relationship $m = -2.5 \log_{10}(B/B_0)$. We want to find the difference $m_B - m_A$. Let $B_A$ and $B_B$ be the brightnesses of Star A and Star B, respectively. We have: $m_B = -2.5 \log_{10}(B_B/B_0)$ and $m_A = -2.5 \log_{10}(B_A/B_0)$. The difference is: $m_B - m_A = [-2.5 \log_{10}(B_B/B_0)] - [-2.5 \log_{10}(B_A/B_0)]$. Factoring out -2.5 gives: $m_B - m_A = -2.5 [\log_{10}(B_B/B_0) - \log_{10}(B_A/B_0)]$. Using the logarithm quotient rule, $\log(x) - \log(y) = \log(x/y)$: $m_B - m_A = -2.5 \log_{10}((B_B/B_0) / (B_A/B_0)) = -2.5 \log_{10}(B_B/B_A)$. We are given that Star A is 100 times brighter than Star B, so $B_A = 100 B_B$, which means $B_B/B_A = 1/100 = 10^{-2}$. Substituting this into the equation: $m_B - m_A = -2.5 \log_{10}(10^{-2})$. Since $\log_{10}(10^{-2}) = -2$, the difference is $m_B - m_A = -2.5 \times (-2) = 5.0$.
2. A power law model $y = ax^b$ is linearized by plotting $Y$ against $X$, where $Y=\ln(y)$ and $X=\ln(x)$. The equation of the resulting regression line is $Y = mX + c$. Which expression correctly gives the value of the coefficient $a$ from the original model?
   Answer: $a=e^c$ The power law model is $y = ax^b$. To linearize it, we take the natural logarithm of both sides: $\ln(y) = \ln(ax^b)$. Using logarithm properties, this becomes $\ln(y) = \ln(a) + \ln(x^b)$, which simplifies to $\ln(y) = \ln(a) + b\ln(x)$. This equation has the form of a straight line, $Y = c + mX$, where $Y=\ln(y)$, $X=\ln(x)$. By comparing the terms, we see that the y-intercept $c$ corresponds to $\ln(a)$ and the gradient $m$ corresponds to $b$. To find $a$ from $c = \ln(a)$, we need to apply the inverse operation of the natural logarithm, which is exponentiation with base $e$. Therefore, $a = e^c$.
3. A student is modelling a relationship with the function $y = 4 \cdot (1.5)^x$. To linearize the data, they decide to plot $\ln(y)$ against $x$. What is the theoretical gradient of this linear graph, correct to 3 significant figures?
   Answer: 0.405 The function is an exponential model of the form $y=ab^x$, with $a=4$ and $b=1.5$. To linearize this model, we take the natural logarithm of both sides: $\ln(y) = \ln(4 \cdot (1.5)^x)$. Using the product rule for logarithms, this becomes $\ln(y) = \ln(4) + \ln((1.5)^x)$. Using the power rule for logarithms, we get $\ln(y) = \ln(4) + x\ln(1.5)$. This equation is in the form of a straight line, $Y = c + mx$, where $Y=\ln(y)$. The y-intercept is $c = \ln(4)$ and the gradient is $m = \ln(1.5)$. Calculating the gradient: $m = \ln(1.5) \approx 0.405465...$. Correct to 3 significant figures, the gradient is 0.405.
4. A population $P$ is modelled by the function $P(t) = A \cdot (10^{0.5})^t$, where $t$ is time in years. The data is linearized by plotting $\log_{10}(P)$ versus $t$. What is the gradient of the resulting line?
   Answer: 0.5 The model is $P(t) = A \cdot (10^{0.5})^t$. This can be rewritten using exponent rules as $P(t) = A \cdot 10^{0.5t}$. To linearize this by plotting $\log_{10}(P)$ versus $t$, we take the logarithm base 10 of both sides: $\log_{10}(P) = \log_{10}(A \cdot 10^{0.5t})$. Using the product rule for logarithms: $\log_{10}(P) = \log_{10}(A) + \log_{10}(10^{0.5t})$. Using the property $\log_b(b^x) = x$, this simplifies to $\log_{10}(P) = \log_{10}(A) + 0.5t$. This is a linear equation of the form $Y = c + mt$, where $Y = \log_{10}(P)$, $t$ is the x-variable, the y-intercept is $c = \log_{10}(A)$, and the gradient is $m = 0.5$.
5. The relationship between the frequency ($f$, in Hz) and tension ($T$, in Newtons) of a vibrating string is modelled by a power law, $f = aT^b$. A log-log plot of $\ln(f)$ against $\ln(T)$ for tensions between 10 N and 100 N yields the regression line $Y = 0.5X + 1.2$. Using this model, what is the predicted frequency for a tension of 200 N, and what can be said about the reliability of this prediction?
   Answer: 46.9 Hz; the prediction is unreliable because it involves extrapolation. First, we predict the frequency. The regression line equation is $\ln(f) = 0.5\ln(T) + 1.2$. For a tension of $T = 200$ N, we calculate $\ln(200) \approx 5.298$. Substituting this into the equation: $\ln(f) = 0.5(5.298) + 1.2 = 2.649 + 1.2 = 3.849$. To find $f$, we compute $f = e^{3.849} \approx 46.94$ Hz. So the predicted frequency is approximately 46.9 Hz. Next, we assess the reliability. The model was created using data for tensions between 10 N and 100 N. The prediction is for a tension of 200 N, which is outside this range. Making predictions outside the range of the original data is called extrapolation. Extrapolated predictions are generally considered unreliable because there is no guarantee that the model holds true for values beyond the observed data range.
6. A dataset is modelled by the exponential function $f(x) = ab^x$. The data is linearized by plotting $\ln(y)$ against $x$. The resulting line passes through the points $(2, 5.5)$ and $(6, 8.3)$. Find the value of $b$ in the model, correct to two decimal places.
   Answer: 2.01 The exponential model is $y = ab^x$. Linearizing by taking the natural logarithm gives $\ln(y) = \ln(a) + x\ln(b)$. This is a linear equation of the form $Y = c + mx$, where $Y = \ln(y)$, the y-intercept $c = \ln(a)$, and the gradient $m = \ln(b)$. The gradient of the line passing through $(x_1, Y_1) = (2, 5.5)$ and $(x_2, Y_2) = (6, 8.3)$ is $m = \frac{Y_2 - Y_1}{x_2 - x_1} = \frac{8.3 - 5.5}{6 - 2} = \frac{2.8}{4} = 0.7$. Since the gradient $m = \ln(b)$, we have $\ln(b) = 0.7$. To find $b$, we calculate $b = e^{0.7} \approx 2.01375...$. Correct to two decimal places, $b = 2.01$.
7. The relationship between two variables, $V$ and $p$, is believed to be $V = c p^{-1.4}$, where $c$ is a constant. If $\log_{10}(V)$ is plotted on the vertical axis and $\log_{10}(p)$ on the horizontal axis, what is the expected gradient of the resulting straight line?
   Answer: -1.4 The given relationship is a power law: $V = c p^{-1.4}$. To linearize this relationship, we take the logarithm of both sides. Using base 10: $\log_{10}(V) = \log_{10}(c p^{-1.4})$. Using logarithm properties, we can expand this to: $\log_{10}(V) = \log_{10}(c) + \log_{10}(p^{-1.4})$. Finally, using the power rule for logarithms: $\log_{10}(V) = \log_{10}(c) - 1.4 \log_{10}(p)$. This equation is in the form $Y = C + mX$, where $Y = \log_{10}(V)$, $X = \log_{10}(p)$, and the y-intercept is $C = \log_{10}(c)$. The gradient, $m$, is the coefficient of the $X$ term, which is -1.4.
8. A scientist proposes a power law model $S = aT^b$ to relate the number of species $S$ on an island to its area $T$ in square kilometers. By plotting $\ln(S)$ versus $\ln(T)$, she obtains the regression line $Y = 0.25X + 1.8$, where $Y=\ln(S)$ and $X=\ln(T)$. Predict the number of species on an island with an area of 200 square kilometers, rounding your answer to the nearest whole number.
   Answer: 23 The linearized equation is $\ln(S) = 0.25\ln(T) + 1.8$. We want to predict $S$ for $T = 200$. First, we calculate $X = \ln(T) = \ln(200) \approx 5.2983$. Next, we substitute this into the regression equation to find $Y = \ln(S)$: $\ln(S) = 0.25 \times 5.2983 + 1.8 \approx 1.3246 + 1.8 = 3.1246$. To find $S$, we take the exponent of this value: $S = e^{3.1246} \approx 22.748$. Rounded to the nearest whole number, the predicted number of species is 23.
9. 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?
   Answer: 93.0 dB The sound level $L$ in decibels (dB) is given by $L = 10 \log_{10}(I/I_0)$, where $I$ is the sound intensity and $I_0$ is a reference intensity. Let $I_1$ be the intensity of one machine. We are given that its sound level is $90 = 10 \log_{10}(I_1/I_0)$. When the second identical machine is turned on, the new intensity is $I_2 = 2I_1$. The new sound level $L_2$ is $L_2 = 10 \log_{10}(2I_1/I_0)$. Using the logarithm property $\log(ab) = \log(a) + \log(b)$, we have $L_2 = 10 (\log_{10}(2) + \log_{10}(I_1/I_0)) = 10\log_{10}(2) + 10\log_{10}(I_1/I_0)$. Since $10\log_{10}(I_1/I_0) = 90$, the equation becomes $L_2 = 10\log_{10}(2) + 90$. Calculating this gives $L_2 \approx 10(0.30103) + 90 = 3.0103 + 90 = 93.0103$. To one decimal place, the new sound level is 93.0 dB.
10. A biologist models the growth of a yeast culture using the equation $P(t) = A \cdot B^t$, where $P$ is the population at time $t$ hours. To linearize the data, the biologist plots $\ln(P)$ against $t$. The resulting line of best fit is given by the equation $Y = 0.15t + 4.20$, where $Y = \ln(P)$. What is the value of $A$ in the biologist's model, correct to the nearest integer?
    Answer: 67 The original model is $P(t) = A \cdot B^t$. Taking the natural logarithm of both sides gives $\ln(P) = \ln(A \cdot B^t) = \ln(A) + \ln(B^t) = \ln(A) + t \ln(B)$. This equation is in the form $Y = mt + c$, where $Y = \ln(P)$, the gradient $m = \ln(B)$, and the y-intercept $c = \ln(A)$. The given line of best fit is $Y = 0.15t + 4.20$. By comparing the two forms, we can see that the y-intercept $\ln(A) = 4.20$. To find $A$, we solve for $A = e^{4.20} \approx 66.686$. Rounded to the nearest integer, $A = 67$.
11. A scientist collects data on a variable $y$ that depends on $x$. To determine the nature of the relationship, two transformations are performed. A plot of $\ln(y)$ versus $x$ has a Pearson's correlation coefficient of $r = 0.88$. A plot of $\ln(y)$ versus $\ln(x)$ has a Pearson's correlation coefficient of $r = -0.99$. Which of the following is the most appropriate conclusion?
    Answer: A power law model of the form $y=ax^b$ with $b<0$ is the best fit for the data. The Pearson's correlation coefficient, $r$, measures the strength and direction of a linear relationship. A value closer to $1$ or $-1$ indicates a stronger linear fit. Here, $r = -0.99$ is much closer to $-1$ than $r = 0.88$ is to $1$. This means the second transformation provides a better linear model for the data. The transformation of plotting $\ln(y)$ versus $\ln(x)$ linearizes a power law relationship of the form $y=ax^b$. The gradient of the line $\ln(y) = b\ln(x) + \ln(a)$ is $b$. Since the correlation coefficient $r$ is negative, the slope of the regression line must be negative, which implies that the exponent $b$ is negative. Thus, a power law model with $b<0$ is the best fit.
12. 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?
    Answer: 5.2 The Richter scale magnitude $M$ is defined as $M = \log_{10}(I/S)$, where $I$ is the intensity of the earthquake and $S$ is a standard intensity. For region A, $M_A = 6.8 = \log_{10}(I_A/S)$. For region B, the intensity is $I_B = I_A / 40$. The magnitude in region B is $M_B = \log_{10}(I_B/S) = \log_{10}((I_A/40)/S)$. Using the logarithm property $\log(x/y) = \log(x) - \log(y)$, we can write $M_B = \log_{10}(I_A/S) - \log_{10}(40)$. We know $\log_{10}(I_A/S) = 6.8$, so $M_B = 6.8 - \log_{10}(40) \approx 6.8 - 1.602 = 5.198$. Rounded to one decimal place, the magnitude is 5.2.
13. It is observed for a certain physical phenomenon that when the logarithm of a quantity $P$ is plotted against the logarithm of a quantity $Q$, the result is a straight line with a negative slope. Which statement best describes the relationship between $P$ and $Q$?
    Answer: As $Q$ increases by a constant percentage, $P$ decreases by a constant percentage. A linear relationship between $\log(P)$ and $\log(Q)$ implies a power law relationship between $P$ and $Q$, of the form $P = aQ^b$. A negative slope means that the exponent $b$ 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 $Q$ is multiplied by a factor $k$, then $P$ is multiplied by a factor $k^b$. Since $b$ is negative, if $k>1$ (an increase), then $0 < k^b < 1$ (a decrease). This means that as $Q$ increases by a constant percentage, $P$ decreases by a constant percentage. Option A describes a linear relationship. Option B describes an exponential relationship.
14. The magnitude $M$ of an earthquake is related to its intensity $I$ by the formula $M = \log_{10}(I/S)$, where $S$ is the intensity of a standard earthquake. An earthquake in Japan was 8000 times as intense as the standard earthquake. A smaller earthquake in California was 500 times as intense as the standard earthquake. What is the difference in the magnitudes of the two earthquakes, to one decimal place?
    Answer: 1.2 Let $M_J$ and $I_J$ be the magnitude and intensity for the Japan earthquake, and $M_C$ and $I_C$ for California. We are given $I_J = 8000S$ and $I_C = 500S$. The difference in magnitudes is $M_J - M_C$. Using the formula: $M_J - M_C = \log_{10}(I_J/S) - \log_{10}(I_C/S)$. Using the logarithm property $\log(a) - \log(b) = \log(a/b)$, we get: $M_J - M_C = \log_{10}((I_J/S) / (I_C/S)) = \log_{10}(I_J / I_C)$. Substituting the given values: $M_J - M_C = \log_{10}(8000S / 500S) = \log_{10}(16)$. Calculating the value: $\log_{10}(16) \approx 1.204$. To one decimal place, the difference is 1.2.
15. A set of bivariate data $(x, y)$ is believed to follow a power law relationship of the form $y=ax^b$. Which of the following transformations would result in a linear graph, and what would the gradient of that graph represent?
    Answer: Plotting $\log(y)$ against $\log(x)$ gives a linear graph with gradient $b$. For a power law relationship $y=ax^b$, 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 $\log(y) = \log(ax^b) = \log(a) + \log(x^b) = \log(a) + b\log(x)$. This equation is in the form of a straight line, $Y = c + mX$, where $Y = \log(y)$, $X = \log(x)$, the y-intercept is $c = \log(a)$, and the gradient is $m = b$. Therefore, plotting $\log(y)$ against $\log(x)$ results in a linear graph with a gradient of $b$.
16. The relationship between the mass $M$ (in kg) and the wingspan $W$ (in m) of a species of bird is modelled by the equation $M = aW^b$. A researcher plots $\log_{10}(M)$ against $\log_{10}(W)$ and obtains a line of best fit with the equation $Y = 2.5X - 0.7$, where $Y = \log_{10}(M)$ and $X = \log_{10}(W)$. Find the value of $a$, correct to two decimal places.
    Answer: 0.20 The original model is the power law $M = aW^b$. Taking the common logarithm (base 10) of both sides gives $\log_{10}(M) = \log_{10}(aW^b) = \log_{10}(a) + \log_{10}(W^b) = \log_{10}(a) + b\log_{10}(W)$. This equation has the linear form $Y = c + mX$, where $Y = \log_{10}(M)$, $X = \log_{10}(W)$, the gradient $m = b$, and the y-intercept $c = \log_{10}(a)$. The given regression line is $Y = 2.5X - 0.7$. Comparing these, we find that $b=2.5$ and the y-intercept $\log_{10}(a) = -0.7$. To find $a$, we calculate $a = 10^{-0.7} \approx 0.1995...$. Rounded to two decimal places, $a = 0.20$.
17. 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?
    Answer: To effectively represent and compare values that span many orders of magnitude on a single, comprehensible scale. 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.
18. The growth of a bacterial colony is modelled by the exponential function $P = P_0 e^{kt}$, where $P$ is the population at time $t$ hours. To analyse the growth rate, a graph of $\ln(P)$ is plotted against $t$. The regression line has a gradient of 0.05. What is the correct interpretation of this gradient?
    Answer: The population grows at a continuous rate of 5% per hour. The model is $P = P_0 e^{kt}$. To linearize this, we take the natural logarithm of both sides: $\ln(P) = \ln(P_0 e^{kt}) = \ln(P_0) + \ln(e^{kt}) = \ln(P_0) + kt$. This is a linear equation in the form $Y = c + mt$, where $Y = \ln(P)$, the y-intercept $c = \ln(P_0)$, and the gradient $m = k$. The given gradient is 0.05, so $k=0.05$. In the model $P = P_0 e^{kt}$, the parameter $k$ represents the continuous relative growth rate. A value of $k=0.05$ corresponds to a continuous growth rate of 5% per hour.
19. A chemist has a solution with a pH of 5. She adds a substance that causes the hydrogen ion concentration, $[H^+]$, to decrease by a factor of 1000. What is the new pH of the solution?
    Answer: 8 The pH of a solution is defined as $pH = -\log_{10}[H^+]$. Let the initial pH be $pH_1 = 5$ and the initial concentration be $[H^+_1]$. Let the final pH be $pH_2$ and the final concentration be $[H^+_2]$. We have $[H^+_2] = [H^+_1] / 1000$. The change in pH is $\Delta pH = pH_2 - pH_1 = -\log_{10}[H^+_2] - (-\log_{10}[H^+_1]) = \log_{10}[H^+_1] - \log_{10}[H^+_2] = \log_{10}([H^+_1]/[H^+_2])$. Since $[H^+_1]/[H^+_2] = 1000$, the change in pH is $\log_{10}(1000) = 3$. So, $pH_2 - pH_1 = 3$. Since $pH_1 = 5$, the new pH is $pH_2 = 5 + 3 = 8$. A decrease in $[H^+]$ (less acidic) corresponds to an increase in pH.
20. The relationship between two variables $x$ and $y$ is given by the power model $y=ax^k$. When $\log_{10}(y)$ is plotted against $\log_{10}(x)$, a straight line is formed which passes through the points $(0.5, 2.1)$ and $(1.5, 5.1)$. Find the value of the exponent $k$.
    Answer: 3.00 The power model is $y=ax^k$. Taking the logarithm base 10 of both sides gives $\log_{10}(y) = \log_{10}(a) + k\log_{10}(x)$. This is a linear equation of the form $Y = c + mX$, where $Y = \log_{10}(y)$, $X = \log_{10}(x)$, the y-intercept $c = \log_{10}(a)$, and the gradient $m = k$. We are given two points on this line: $(X_1, Y_1)=(0.5, 2.1)$ and $(X_2, Y_2)=(1.5, 5.1)$. The gradient of this line is $k = m = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{5.1 - 2.1}{1.5 - 0.5} = \frac{3.0}{1.0} = 3$. Thus, the value of the exponent $k$ is 3.00.

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