# Equations Involving Exponentials

## Equations Involving Exponentials Revision

**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**.

The following topics build on the content in this page.

**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 =

**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

**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}

**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}

## Equations Involving Exponentials Example Questions

**Question 1: **Solve for x:

**[3 marks]**

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

**Question 2: **Solve for x:

**[5 marks]**

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.

**Question 3: **A particle’s radioactivity decreases according to R=1000\times0.8^{T} where R is radioactivity and T is time in years.

a) Find the half-life (when radioactivity has halved from the initial value) of the particle.

b) A safe level of radioactivity is 5. How long until the particle is safe?

**[5 marks]**

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}