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: $3times6^{2x}-7times6^{x}+2=0$

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

$y=6^{x}$

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

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

$y=dfrac{1}{3}$ 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: $log_{3}(21x-2)-2log_{3}(x)=3$

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

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

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

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

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

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

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

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

$x=dfrac{1}{9}$ 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 $V=14000times0.9^{T}$ 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 $V=14000times0.9^{0}=14000$

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

iii) She paid new cost which is $14000$

Half of what she paid is $7000$

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

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

$0.9^{T}=0.5$

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