The function \( y(t) \) is given by \( y=\mathrm{e}^{t} \) (a) Calculate values of \( t \) such that \( y \) is (i) 1 , (ii) 2 , (iii) 4 , (iv) 8 . (b) Show that if \( t \) is increased by \( \ln 2 \) then the value of \( y \) doubles. (c) A function \( z(t) \) is given by

To find the values of \( t \) such that \( y = \mathrm{e}^{t} \): (i) For \( y = 1 \): We set \( \mathrm{e}^{t} = 1 \). Taking the natural logarithm on both sides gives \( t = \ln(1) = 0 \). (ii) For \( y = 2 \): Setting \( \mathrm{e}^{t} = 2 \), we find \( t = \ln(2) \). (iii) For \( y = 4 \): Here, \( \mathrm{e}^{t} = 4 \), leading to \( t = \ln(4) = 2 \ln(2) \). (iv) For \( y = 8 \): We set \( \mathrm{e}^{t} = 8 \), resulting in \( t = \ln(8) = 3 \ln(2) \). Now for part (b): To show that if \( t \) is increased by \( \ln 2 \) then the value of \( y \) doubles, we calculate \( y(t + \ln(2)) \): \[ y(t + \ln(2)) = \mathrm{e}^{t + \ln(2)} = \mathrm{e}^{t} \cdot \mathrm{e}^{\ln(2)} = \mathrm{e}^{t} \cdot 2 = 2y(t) \] Thus, indeed \( y \) doubles when \( t \) is increased by \( \ln(2) \). For part (c), if you wish to introduce \( z(t) \), you might consider a function defined similarly, such as \( z(t) = \mathrm{e}^{kt} \) where \( k \) is a constant. This form can model exponential growth or decay, depending on whether \( k \) is positive or negative, providing a versatile approach to study varying phenomena in the real world, from populations to radioactive decay.