y = kxⁿ: lg y against lg x questions with answers

Questions

1. Question 1The power n from the gradient

   The variables $x$ and $y$ are related by the equation $y = kx^{n}$, where $k$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$, a straight line passing through the points $(0.3, 0.18)$ and $(1.1, -0.62)$ is obtained. Find the value of $n$.

   Show the worked solution

   1. Step 1: Take lg of both sides and use the laws of logarithms$\lg y = \lg k + n\lg x$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = lg x$\lg y = n\lg x + \lg k$
   3. Step 3: Compare with Y = mX + c$n$ 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$n = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$n = \dfrac{-0.62 - 0.18}{1.1 - 0.3}$$= -1$
2. Question 2The power n from the gradient

   The variables $x$ and $y$ are related by the equation $y = kx^{n}$, where $k$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$, a straight line passing through the points $(0.3, 1.87)$ and $(1.1, 3.87)$ is obtained. Find the value of $n$.

   Show the worked solution

   1. Step 1: Take lg of both sides and use the laws of logarithms$\lg y = \lg k + n\lg x$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = lg x$\lg y = n\lg x + \lg k$
   3. Step 3: Compare with Y = mX + c$n$ 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$n = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$n = \dfrac{3.87 - 1.87}{1.1 - 0.3}$$= 2.5$
3. Question 3k from the intercept

   The variables $x$ and $y$ are related by the equation $y = kx^{n}$, where $k$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$, a straight line passing through the points $(0.1, 0.68)$ and $(0.9, 2.28)$ 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$\lg y = \lg k + n\lg x$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = lg x$\lg y = n\lg x + \lg k$
   3. Step 3: Compare with Y = mX + c$n$ 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$n = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$n = \dfrac{2.28 - 0.68}{0.9 - 0.1}$$= 2$
   5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c$0.68 = 2(0.1) + \lg k$$\lg k = 0.48$
   6. Step 6: Undo the lg: if lg k = c, then k = 10 to the power c$k = 10^{0.48}$$= 3.02$ (3 s.f.)
4. Question 4k from the intercept

   The variables $x$ and $y$ are related by the equation $y = kx^{n}$, where $k$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$, a straight line passing through the points $(0.5, 1.85)$ and $(1.5, 3.85)$ 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$\lg y = \lg k + n\lg x$
   2. Step 2: Write it as Y = mX + c, with Y = lg y and X = lg x$\lg y = n\lg x + \lg k$
   3. Step 3: Compare with Y = mX + c$n$ 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$n = \dfrac{Y_2 - Y_1}{X_2 - X_1}$$n = \dfrac{3.85 - 1.85}{1.5 - 0.5}$$= 2$
   5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c$1.85 = 2(0.5) + \lg k$$\lg k = 0.85$
   6. Step 6: Undo the lg: if lg k = c, then k = 10 to the power c$k = 10^{0.85}$$= 7.08$ (3 s.f.)