The exponential function e^xEdexcel International A Level Maths: Revision notes

The number $e\approx2.718$ is the base for which $y=e^x$ has gradient equal to its own value at every point. The graph of $y=e^x$ passes through $(0,1)$, is always above the $x$-axis ($e^x>0$ for all $x$), increases for all $x$, and has the $x$-axis as a horizontal asymptote: $e^x\to0$ as $x\to-\infty$ and $e^x\to\infty$ as $x\to\infty$. Its range is $y>0$. The graph of $y=e^{-x}$ is its reflection in the $y$-axis, a decaying curve through $(0,1)$.

Start from $y=e^x$ and apply transformations in a sensible order. For $y=e^{ax+b}+c$:

• $+c$ is a translation of $\begin{pmatrix}0\\ c\end{pmatrix}$, so the asymptote moves from $y=0$ to $y=c$ and the range becomes $y>c$.
• $e^{x+b}$ is a translation of $\begin{pmatrix}-b\\ 0\end{pmatrix}$ and $e^{ax}$ is a horizontal stretch with scale factor $\frac1a$. Neither changes the asymptote.
• The $y$-intercept is found by putting $x=0$: $e^{b}+c$. Example: $y=e^{2x-1}+3$ has asymptote $y=3$, range $y>3$ and $y$-intercept $(0,\,e^{-1}+3)$. A multiplier in front, as in $y=80e^{-0.05t}$, stretches the curve vertically and gives the value at $t=0$ directly.

To undo an exponential, take natural logarithms of both sides. Since $\ln(e^u)=u$: $e^{ax+b}=p\ \Rightarrow\ ax+b=\ln p\ \Rightarrow\ x=\frac{\ln p-b}{a}.$ This needs $p>0$; if $p\le0$ there is no solution. Leave answers in exact form (such as $x=\frac13\ln25$) when asked, and round only at the end. Worked example: solve $e^{3x}-5=20$. Rearrange to $e^{3x}=25$, so $3x=\ln25$ and $x=\frac13\ln25=1.07$ (3 s.f.).

Models of growth and decay use $y=Ae^{kt}$ (growth when $k>0$) or $y=Ae^{-kt}$ (decay). The constant $A$ is the starting value at $t=0$. Models of cooling use $\theta=\theta_0+Ae^{-kt}$, where the temperature approaches the constant $\theta_0$ (the asymptote) in the long term. To find a constant such as $k$, substitute known values and solve the resulting equation of the form $e^{ax+b}=p$ using logarithms. To find a time, substitute the target value, isolate the exponential, then take $\ln$. Worked example: $80e^{-0.05t}=20\Rightarrow e^{-0.05t}=0.25\Rightarrow t=\frac{\ln4}{0.05}=27.7$.