# The Exponential Function

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## The Exponential Function

We have met exponential functions before, but there is one specific exponential function that has special properties, and it is based around a special number: $\color{red}e$.

A Level    ## e

The exponential function is $\color{red}e^{x}$.

$\color{red}e=2.71828...$ is a number. It is a decimal that goes on forever
(like $\pi$).

$\color{red}e^{x}$ has special properties, most notable being that the gradient of $\color{red}e^{x}$ is $\color{red}e^{x}$. This will be very important in the differentiation section of the course.

There are some key facts to remember about the graph of $y=e^{x}$:

• It crosses the $y$-axis at $(0,1)$
• As $x\rightarrow\infty$, $\color{red}e^{x}\color{grey}\rightarrow\infty$ and as $x\rightarrow -\infty$, $\color{red}e^{x}\color{grey}\rightarrow0$
• $\color{red}e^{x}$ is never negative.

$y=e^{ax+b} + c$ is a transformation of $y = e^x$, where $a$ is a horizontal stretch, $b$ moves it horizontally and $c$ moves it vertically.

$y = e^{-x}$ reflects $y=e^x$ in the $y$-axis.

A Level    ## Natural Logarithm

The inverse function of $\color{red}e^{x}$ is the natural logarithm $\color{blue}\ln(x)$. This is the logarithm with base $\color{red}e$ $(\text{log}_e (x))$.

All the laws of logarithms can be applied to the natural logarithm.

$\color{blue}\ln(a)\color{grey}+\color{blue}\ln(b)\color{grey}=\color{blue}\ln(ab)$

$\color{blue}\ln(a)\color{grey}-\color{blue}\ln(b)\color{grey}=\color{blue}\ln\left(\dfrac{a}{b}\right)$

$\color{blue}\ln(a^{b})\color{grey}=\color{blue}b\ln(a)$

The graph of the natural logarithm (in blue) is the reflection in the line $y=x$ of the graph of the exponential function (in red).

There are key facts to remember about the graph of $\color{blue}y=\ln(x)$:

• It crosses the $x$-axis at $(1,0)$
• As $x\rightarrow\infty$, $\color{blue}\ln(x)\color{grey}\rightarrow\infty$ and as $x\rightarrow0$, $\color{blue}\ln(x)\color{grey}\rightarrow -\infty$
• $\color{blue}\ln(x)$ does not take any values for $x\leq0$

Since $\ln (x)$ is the inverse of $e^x$ and is a logarithmic function, we have these formulas relating the two:

$\textcolor{red}{e}^{\textcolor{blue}{\ln (x)}} = x$

$\textcolor{blue}{\ln} \textcolor{red}{(e^x)} = x$

A Level   A Level   ## Example 1: Equations Involving the Exponential Function

Solve for $x$:

$\color{red}e^{3x}\color{grey}=10$

[2 marks]

$\color{red}e^{3x}\color{grey}=10$

$3x=\color{blue}\ln(10)$

\begin{aligned}x&=\dfrac{\color{blue}{\ln(10)}}{3}\\[1.2em]&=0.768\end{aligned}

A Level   ## Example 2: Equations Involving Logarithms

Solve for $x$:

$\color{blue}\ln(4x+3)\color{grey}=2$

[2 marks]

$\color{blue}\ln(4x+3)\color{grey}=2$

$4x+3=\color{red}{e^{2}}$

$4x=\color{red}e^{2}\color{grey}-3$

\begin{aligned}x&=\dfrac{1}{4}(\color{red}e^{2}\color{grey}-3)\\[1.2em]&=1.10\end{aligned}

A Level   ## Example Questions

a) $e^{x}=2$

\begin{aligned}x&=\ln(2)\\[1.2em]&=0.693\end{aligned}

b) $e^{5x}=19$

$5x=\ln(19)$

\begin{aligned}x&=\dfrac{1}{5}\ln(19)\\[1.2em]&=0.589\end{aligned}

c) $e^{12x}=234$

$12x=\ln(234)$

\begin{aligned}x&=\dfrac{1}{12}\ln(234)\\[1.2em]&=0.455\end{aligned}

a) $\ln(x+1)=4$

$x+1=e^{4}$

\begin{aligned}x&=e^{4}-1\\[1.2em]&=53.6\end{aligned}

b) $\ln(3x+2)=1.5$

$3x+2=e^{1.5}$

$3x=e^{1.5}-2$

\begin{aligned}x&=\dfrac{1}{3}(e^{1.5}-2)\\[1.2em]&=0.827\end{aligned}

c) $\ln(9x+36)=0.6$

$9x+36=e^{0.6}$

$9x=e^{0.6}-36$

\begin{aligned}x&=\dfrac{1}{9}(e^{0.6}-36)\\[1.2em]&=-3.80\end{aligned}

$e^{2x}-13e^{x}+36=0$

Note that $e^{2x}=(e^{x})^{2}$

$(e^{x})^{2}-13e^{x}+36=0$

Substitute: $y=e^{x}$

$y^{2}-13y+36=0$

$(y-9)(y-4)=0$

$y=9$ or $y=4$

Reverse substitution:

$e^{x}=9$ or $e^{x}=4$

$x=\ln(9)$ or $x=\ln(4)$

$x=2.20$ or $x=1.39$

$\ln(4x+3)-2\ln(x)=5$

$\ln(4x+3)-\ln(x^{2})=5$

$\ln\left(\dfrac{4x+3}{x^{2}}\right)=5$

$\dfrac{4x+3}{x^{2}}=e^{5}$

$4x+3=e^{5}x^{2}$

$e^{5}x^{2}-4x-3=0$

\begin{aligned}x&=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\[1.2em]&=\dfrac{4\pm\sqrt{(-4)^{2}-4\times e^{5}\times(-3)}}{2e^{5}}\\[1.2em]&=\dfrac{4\pm\sqrt{16+12e^{5}}}{2e^{5}}\\[1.2em]&=\dfrac{2\pm\sqrt{4+3e^{5}}}{e^{5}}\end{aligned}

$x=0.156$ or $x=-0.129$

We can discount the negative solution because $\ln(x)$ is not valid for negative $x$.

$x=0.156$

A Level
A Level

A Level

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