Equations Involving Exponentials

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}