Exponential laws: lg y or ln y against x questions with answers

Questions

1. Question 1y = kbˣ

   The variables $x$ and $y$ are related by the equation $y = kb^{x}$, where $k$ and $b$ are constants. When $\lg y$ is plotted against $x$, a straight line passing through the points $(2, 1.15)$ and $(6, 2.15)$ is obtained. Find the value of $k$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Take lg of both sides and use the laws of logarithms$\lg y = \lg k + x\lg b$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = x$\lg y = (\lg b)x + \lg k$
   3. Step 3: Compare with Y = mX + c$\lg b$ is the gradient and $\lg k$ is the $Y$-intercept.
   4. Step 4: Gradient of the line through the two points: put them into the gradient formula$\lg b = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$\lg b = \dfrac{2.15 - 1.15}{6 - 2}$$= 0.25$
   5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c$1.15 = 0.25(2) + \lg k$$\lg k = 0.65$
   6. Step 6: Undo the lg: if lg k = c, then k = 10 to the power c$k = 10^{0.65}$$= 4.47$ (3 s.f.)
2. Question 2y = kbˣ

   The variables $x$ and $y$ are related by the equation $y = kb^{x}$, where $k$ and $b$ are constants. When $\lg y$ is plotted against $x$, a straight line passing through the points $(2, 2.3)$ and $(6, 4.7)$ is obtained. Find the value of $k$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Take lg of both sides and use the laws of logarithms$\lg y = \lg k + x\lg b$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = x$\lg y = (\lg b)x + \lg k$
   3. Step 3: Compare with Y = mX + c$\lg b$ is the gradient and $\lg k$ is the $Y$-intercept.
   4. Step 4: Gradient of the line through the two points: put them into the gradient formula$\lg b = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$\lg b = \dfrac{4.7 - 2.3}{6 - 2}$$= 0.6$
   5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c$2.3 = 0.6(2) + \lg k$$\lg k = 1.1$
   6. Step 6: Undo the lg: if lg k = c, then k = 10 to the power c$k = 10^{1.1}$$= 12.6$ (3 s.f.)
3. Question 3y = aeᵇˣ

   The variables $x$ and $y$ are related by the equation $y = ae^{bx}$, where $a$ and $b$ are constants. When $\ln y$ is plotted against $x$, a straight line passing through the points $(1, 1.6)$ and $(4, 3.1)$ is obtained. Find the value of $b$.

   Show the worked solution

   1. Step 1: Take ln of both sides and use the laws of logarithms$\ln y = \ln a + bx$
   2. Step 2: Write it as Y = mX + c, with Y = ln y and X = x$\ln y = bx + \ln a$
   3. Step 3: Compare with Y = mX + c$b$ is the gradient and $\ln a$ is the $Y$-intercept.
   4. Step 4: Gradient of the line through the two points: put them into the gradient formula$b = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$b = \dfrac{3.1 - 1.6}{4 - 1}$$= 0.5$
4. Question 4y = aeᵇˣ

   The variables $x$ and $y$ are related by the equation $y = ae^{bx}$, where $a$ and $b$ are constants. When $\ln y$ is plotted against $x$, a straight line passing through the points $(1, 1.9)$ and $(3, 3.5)$ is obtained. Find the value of $b$.

   Show the worked solution

   1. Step 1: Take ln of both sides and use the laws of logarithms$\ln y = \ln a + bx$
   2. Step 2: Write it as Y = mX + c, with Y = ln y and X = x$\ln y = bx + \ln a$
   3. Step 3: Compare with Y = mX + c$b$ is the gradient and $\ln a$ is the $Y$-intercept.
   4. Step 4: Gradient of the line through the two points: put them into the gradient formula$b = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$b = \dfrac{3.5 - 1.9}{3 - 1}$$= 0.8$