14) Use logarithms to express the following equations in the form \( y=m x+c \) and state the gradients of each line (a) \( P=6 x^{5} \) (b) \( y=5 e^{3 x} \)

To express the given equations in the linear form \( y = m x + c \) using logarithms and determine the gradients (\( m \)), follow the steps below for each part: --- ### **(a) \( P = 6 x^{5} \)** **Step 1: Take the natural logarithm of both sides.** \[ \ln P = \ln(6 x^{5}) \] **Step 2: Use logarithm properties to simplify.** \[ \ln P = \ln 6 + \ln x^{5} = \ln 6 + 5 \ln x \] **Step 3: Rearrange to match the linear form \( Y = m X + c \).** \[ \ln P = 5 \ln x + \ln 6 \] Here, if we let: - \( Y = \ln P \) - \( X = \ln x \) Then the equation becomes: \[ Y = 5X + \ln 6 \] **Gradient:** \[ m = 5 \] --- ### **(b) \( y = 5 e^{3 x} \)** **Step 1: Take the natural logarithm of both sides.** \[ \ln y = \ln(5 e^{3x}) \] **Step 2: Use logarithm properties to simplify.** \[ \ln y = \ln 5 + \ln e^{3x} = \ln 5 + 3x \] **Step 3: Rearrange to match the linear form \( Y = m X + c \).** \[ \ln y = 3x + \ln 5 \] Here, if we let: - \( Y = \ln y \) - \( X = x \) Then the equation becomes: \[ Y = 3X + \ln 5 \] **Gradient:** \[ m = 3 \] --- ### **Summary of Results** - **(a)** After taking logarithms, the equation becomes: \[ \ln P = 5 \ln x + \ln 6 \] **Gradient:** \( m = 5 \) - **(b)** After taking logarithms, the equation becomes: \[ \ln y = 3x + \ln 5 \] **Gradient:** \( m = 3 \) --- These linear forms allow you to plot \( \ln P \) versus \( \ln x \) for part (a) and \( \ln y \) versus \( x \) for part (b), with the gradients representing the coefficients of \( \ln x \) and \( x \), respectively.