# Equations Involving Exponentials

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Equations Involving Exponentials

Equations involving exponentials and logarithms can become much more complicated than we have already seen. On this page, we will learn how to use a calculator for logarithms, attempt to solve some more difficult equations, and apply knowledge of logarithms to real life.

A Level   ## Using a Calculator

Your calculator will have several buttons to do with logarithms.

$\log_{\square}\square$ allows you to do any logarithm. Put the base in the lower box and the number in the box on the right.

$\log_{10}\square$ is for logarithms with a base of $10$ only.

$\ln$ is the natural logarithm. We will see this later.

Example: If we want to do $\log_{4}(81)$ on a calculator we would press $\log_{\square}\square$ then $4$ then $8$ then $1$ then $=$

A Level   ## Exponential Equations

Some exponential equations look more complicated, but they are really just quadratics in disguise.

They take the form $ap^{2x}+bp^{x}+c=0$

We solve them by substituting $y=p^{x}$, and noticing that $y^{2}=p^{2x}$

Then we have a simple quadratic: $ay^{2}+by+c=0$

This gives us two roots: $y=r_{1}$ and $y=r_{2}$

Putting $x$ back in gives $p^{x}=r_{1}$ and $p^{x}=r_{2}$, which are two equations that we already know how to solve.

Example: $3\times6^{2x}-7\times6^{x}+2=0$

$3\times(6^{x})^{2}-7\times6^{x}+2=0$

$y=6^{x}$

$3y^{2}-7y-2=0$

$(3y-1)(y-2)=0$

$y=\dfrac{1}{3}$ or $y=2$

$6^{x}=\dfrac{1}{3}$ or $6^{x}=2$

$x=\log_{6}\left(\dfrac{1}{3}\right)$ or $x=\log_{6}(2)$

$x=-0.613$ or $x=0.387$

A Level   ## Logarithm Equations

Difficult logarithm equations require you to use more than one law of logarithms.

Example: $\log_{3}(21x-2)-2\log_{3}(x)=3$

$\log_{3}(21x-2)+\log_{3}(x^{-2})=3$

$\log_{3}((21x-2)x^{-2})=3$

$\log_{3}\left(\dfrac{21x-2}{x^{2}}\right)=3$

$\dfrac{21x-2}{x^{2}}=3^{3}$

$\dfrac{21x-2}{x^{2}}=27$

$21x-2=27x^{2}$

$27x^{2}-21x+2=0$

$(9x-1)(3x-2)=0$

$x=\dfrac{1}{9}$ and $x=\dfrac{2}{3}$

A Level   ## Real-Life Problems

Exponentials and logarithms appear frequently in real life. An important skill is being able to solve real world problems involving exponentials and logarithms.

Example: A car depreciates in value over the course of several years according to the function $V=14000\times0.9^{T}$ where $V$ is the value of the car and $T$ is the time in years.

i) How much does the car cost new?

ii) Jon has owned his car for $9$ years. How much is it worth?

iii) Clarissa buys a new car today. How many years until it is worth half of what she paid?

i) New is at $T=0$ so $V=14000\times0.9^{0}=14000$

ii) This is $T=9$ so $V=14000\times0.9^{9}=5424$

iii) She paid new cost which is $14000$

Half of what she paid is $7000$

$7000=14000\times0.9^{T}$

$0.9^{T}=\dfrac{7000}{14000}$

$0.9^{T}=0.5$

\begin{aligned}T&=\log_{0.9}(0.5)\\[1.2em]&=6.58\end{aligned}

A Level   ## Example Questions

$3^{2x}-28\times3^{x}+27=0$

$(3^{x})^{2}-28\times3^{x}+27=0$

Make a substitution $y=3^{x}$

$y^{2}-28y+27=0$

$(y-27)(y-1)=0$

$y=27$ or $y=1$

Put our substitution back in:

$3^{x}=27$ or $3^{x}=1$

$x=\log_{3}(27)$ or $x=\log_{3}(1)$

$x=3$ or $x=0$

$2\log_{2}(2x-1)-3\log_{2}(x)=4$

$\log_{2}((2x-1)^{2})-\log_{2}(x^{3})=4$

$\log_{2}\left(\dfrac{(2x-1)^{2}}{x^{3}}\right)=4$

$\dfrac{(2x-1)^{2}}{x^{3}}=2^{4}$

$\dfrac{(2x-1)^{2}}{x^{3}}=16$

$(2x-1)^{2}=16x^{3}$

$4x^{2}-4x+1=16x^{3}$

$16x^{3}-4x^{2}+4x-1=0$

$(4x-1)(4x^{2}+1)=0$

$x=\dfrac{1}{4}$ is the only real solution.

$R=1000\times0.8^{T}$

Find initial value, which is at $T=0$:

\begin{aligned}R&=1000\times0.8^{0}\\[1.2em]&=1000\times1\\[1.2em]&=1000\end{aligned}

Hence, half-life is when $R=\dfrac{1000}{2}=500$

$500=1000\times0.8^{T}$

$\dfrac{500}{1000}=0.8^{T}$

$0.5=0.8^{T}$

\begin{aligned}T&=\log_{0.8}(0.5)\\[1.2em]&=3.11\text{ years}\end{aligned}

$R=5$

$5=1000\times0.8^{T}$

$\dfrac{5}{1000}=0.8^{T}$

$0.005=0.8^{T}$

\begin{aligned}T&=\log_{0.8}(0.005)\\[1.2em]&=23.7\text{ years}\end{aligned}

A Level

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