Exponentials and Logarithms

A LevelAQAEdexcelOCRAQA November 2022Edexcel November 2022OCR November 2022

Exponentials and Logarithms

A sum such as $4+3=7$ has two inverses: $7-3=4$ and $7-4=3$.

A product such as $8\times6=48$ has two inverses: $48\div6=8$ and $48\div8=6$.

An exponential (power) such as $3^{4}=81$ has an inverse of the fourth root: $\sqrt[4]{81}=3$. But for the pattern to continue there must be another inverse – an operation involving $81$ and $3$ to get back to $4$.

This other inverse is the logarithm.

A Level

Logarithms are the Inverse of Exponentials

$\log_{a}(b)=c$ means that $a^{c}=b$

$a^{1}=a$ for any $a$ so $\log_{a}(a)=1$

$a^{0}=1$ for any $a$ so $\log_{a}(1)=0$

Example: $\log_{3}(81)=4$ because $3^{4}=81$

We also have other rules:

$\log_{a}(x)+\log_{a}(y)=\log_{a}(xy)$

$\log_{a}(x)-\log_{a}(y)=\log_{a}\left(\dfrac{x}{y}\right)$

$\log_{a}(x^{n})=n\log_{a}(x)$

$\log_{a}(x)=\dfrac{\log_{b}(x)}{\log_{b}(a)}$

Note: The little number after the $\text{log}$ is called the base. The most common base is $10$, but this is usually left out – i.e. for $\text{log}_{10}$ we just write $\text{log}$.

A Level

Using Logarithms and Exponentials to Solve Equations

We can use the rules above – called the laws of logarithms – to solve equations that involve exponentials.

Example: $3^{4x}=5$

$\log(3^{4x})=\log(5)$

$4x\log(3)=\log(5)$

\begin{aligned}x&=\dfrac{\log(5)}{4\log(3)}\\[1.2em]&=0.366\end{aligned}

Note: The base of the logarithm was not given in this example. This is because we would get the same answer using any base.

On the flipside, we can use exponentials to solve equations involving logarithms.

Example: $6\log_{4}(x)=17$

$\log_{4}(x)=\dfrac{17}{6}$

\begin{aligned}x&=4^{\frac{17}{6}}\\[1.2em]&=50.8\end{aligned}

A Level

Graphs of Exponentials

An exponential graph is a graph of the function $y=a^{x}$ for some $a>0$. They all have the same basic shape. If $a>1$ then $y$ increases as $x$ increases. If $a<1$ then $y$ decreases as $x$ increases. Larger $a$ gives faster increase if $a>1$ while smaller $a$ gives faster decrease if $a<1$.

Note: Exponential graphs never reach $\mathbf{0}$.

On the left graph is $y=2^{x}$, $y=3^{x}$ and $y=4^{x}$.

On the right graph is $y=\left(\dfrac{1}{2}\right)^{x}$, $y=\left(\dfrac{1}{3}\right)^{x}$ and $y=\left(\dfrac{1}{4}\right)^{x}$.

A Level

Example Questions

a) $\log_{7}(343)=3$

b) $\log_{5}(625)=4$

c) $\log_{2}(65536)=16$

d) $\log_{729}(9)=\dfrac{1}{3}$

e) $\log_{22}(1)=0$

a) $4$ because $2^{4}=16$

b) $3$ because $5^{3}=125$

c) $5$ because $3^{5}=243$

d) $2$ because $\left(\dfrac{1}{4}\right)^{2}=\dfrac{1}{16}$

a)

\begin{aligned}\log_{a}(12)+\log_{a}(6)&=\log_{a}(12\times6)\\[1.2em]&=\log_{a}(72)\end{aligned}

b)

\begin{aligned}2\log_{b}(5)+\log_{b}(4)&=\log_{b}(5^{2})+\log_{b}(4)\\[1.2em]&=\log_{b}(25)+\log_{b}(4)\\[1.2em]&=\log_{b}(25\times4)\\[1.2em]&=\log_{b}(100)\end{aligned}

c)

\begin{aligned}3\log_{6}(2)+\dfrac{1}{2}\log_{6}(9)&=\log_{6}(2^{3})+\log_{6}(9^{\frac{1}{2}})\\[1.2em]&=\log_{6}(8)+\log_{6}(3)\\[1.2em]&=\log_{6}(8\times3)\\[1.2em]&=\log_{6}(24)\end{aligned}

$6^{7x}=38$

$7x=\log_{6}(38)$

$x=\dfrac{1}{7}\log_{6}(38)$

$x=0.290$

$\log_{6}(x-2)+\log_{6}(x-7)=2$

$\log_{6}((x-2)(x-7))=2$

$(x-2)(x-7)=6^{2}$

$x^{2}-9x+14=36$

$x^{2}-9x-22=0$

$(x-11)(x+2)=0$

$x=11$ or $x=-2$

A Level

A Level

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