Solving Exponential Equations & Inequalities (College Board AP® Precalculus): Study Guide

Exponential equations & inequalities

How do I solve an exponential equation?

• An exponential equation is an equation where the variable appears in an exponent

  • E.g. $3^x=81$ or $5e^{(x+3)}=40$

• The general strategy for solving exponential equations involves isolating the exponential expression

  • and then using logarithms (or properties of exponents) to solve for the variable

• Method 1: Rewriting with a common base

  • If both sides of the equation can be written as powers of the same base, you can set the exponents equal

  • E.g. to solve $2^{(x+1)}=16$

    • Rewrite $16$ as $2^4$

    • So $2^{(x+1)}=2^4$

    • Therefore $x+1=4$, giving $x=3$

  • This method works well when the numbers involved are recognizable powers of a common base

• Method 2: Taking logarithms of both sides

  • When the equation cannot be rewritten with a common base, take the logarithm of both sides

  • You can use any base of logarithm

    • but the natural logarithm ($\ln$) is most common, especially with base $e$ equations

  • E.g. to solve $e^{(x+3)}=10$

    • Take $\ln$ of both sides: $\ln \left(e^{(x+3)}\right)=\ln 10$

    • Simplify the left side: $x+3=\ln 10$

      • $\ln$ cancels $e^{◻}$ because they are inverses

    • Solve: $x=-3+\ln 10$

  • For equations with bases other than $e$, the same approach works

  • E.g. to solve $2.916·(0.7{)}^x=2$

    • Divide both sides by 2.916: $(0.7{)}^x=\frac{2}{2.916}$

    • Take $\ln$ of both sides: $\ln (0.7^x)=\ln \left(\frac{2}{2.916}\right)$

    • Use properties of logarithms: $x\ln (0.7)=\ln \left(\frac{2}{2.916}\right)$

    • Divide: $x=\frac{\ln (2/2.916)}{\ln (0.7)}\approx 1.057$

• Method 3: Using the inverse relationship $b^x=c\iff x={\log }_{b}c$

  • This is equivalent to taking logarithms

    • but uses the definition of a logarithm directly

  • E.g. $e^{(x+3)}=10$ means $x+3={\log }_{e}10=\ln 10$

• The key property that makes all these methods work is the inverse relationship between exponential and logarithmic functions

What about exponential equations that involve additional terms?

• Sometimes you need to isolate the exponential term before applying logarithms

  • E.g. to solve $4e^{(x-2)}+1=5$

    • Subtract 1: $4e^{(x-2)}=4$

    • Divide by 4: $e^{(x-2)}=1$

    • Since $e^0=1$, it follows that $x-2=0$

    • so $x=2$

• In general, for equations of the form $a{e}^{(bx+c)}+d=k$:

  • Subtract $d$: $a{e}^{(bx+c)}=k-d$

  • Divide by $a$: $e^{(bx+c)}=\frac{k-d}{a}$

  • Take $\ln$: $bx+c=\ln \left(\frac{k-d}{a}\right)$

  • Solve for $x$: $x=\frac{\ln \left(\frac{k-d}{a}\right)-c}{b}$

• After isolating the exponential expression, always check whether the right-hand side is positive

  • Since $b^x>0$ for any base $b>0$, an equation like $e^x=-3$ has no solution

  • This connects to checking for extraneous solutions (see below)

What is a 'hidden quadratic' exponential equation?

• Some exponential equations can be transformed into quadratic equations using a substitution

  • This happens when the equation contains terms involving both $b^{2x}$ and $b^x$

  • Since $b^{2x}=(b^x{)}^2$, substituting $y=b^x$ converts the equation into a quadratic in $y$

• E.g. to solve $e^{2x}-5e^x+6=0$:

  • Let $y=e^x$, so $e^{2x}=y^2$

    • The equation becomes $y^2-5y+6=0$

  • Factor: $(y-2\left)\right(y-3)=0$

    • So $y=2$ or $y=3$

  • Back-substitute

    • $e^x=2\Rightarrow x=\ln 2$

    • or $e^x=3\Rightarrow x=\ln 3$

• This technique also works with bases other than $e$

• E.g. $4^x-3·2^x+2=0$

  • can be rewritten as $(2^x{)}^2-3·2^x+2=0$

    • because $4^x={\left(2^2\right)}^x=2^{2x}={\left(2^x\right)}^2$

  • Let $y=2^x$

    • $y^2-3y+2=0$

    • giving $(y-1\left)\right(y-2)=0$

  • So $2^x=1\Rightarrow x=0$

    • or $2^x=2\Rightarrow x=1$

What are extraneous solutions, and why do I need to check for them?

• When solving exponential equations, you should always check whether the solutions you find are valid

• Solutions can be extraneous (invalid) for two reasons:

• Mathematical limitations

  • The range of an exponential function means that $b^x>0$ for all real $x$ (when $b>0$)

  • So if a substitution or manipulation leads to $b^x=\text{negative number}$

    • there is no solution from that part

  • E.g. if solving a hidden quadratic gives $e^x=-5$ as one of the answers

    • this must be rejected because $e^x$ is always positive

  • This is the most common source of extraneous solutions in exponential equations

• Contextual limitations

  • In applied problems, the domain may be restricted

    • E.g. if $t$ represents time and $t\ge 0$, then a negative solution for $t$ would be extraneous

  • Or if a model is only valid for certain input values

    • then solutions outside that set of values should be rejected

How do I solve exponential inequalities?

• Exponential inequalities are solved using similar techniques to exponential equations

• A key principle here is that an exponential function is either always increasing or always decreasing

  • This depends on the value of the base $b$

    • If $\mathit{b}\mathbf{>}\mathbf{1}$, the exponential function $b^x$ is increasing

    • If $\mathbf{0}\mathbf{<}\mathit{b}\mathbf{<}\mathbf{1}$, the exponential function $b^x$ is decreasing

• This means that if $x=p$ is the solution to $b^x=k$

  • then for everywhere on one side of $p$

    • $b^x>k$ is true

  • and for everywhere on the other side of $p$

    • $b^x<k$ is true

  • $b^x$ being increasing or decreasing will tell you which side is which

• E.g. $0.5^x>0.125$

  • Start by solving as an equation $0.5^x=0.125$

    • take ${\log }_{0.5}$ of both sides: ${\log }_{0.5}\left(0.5^x\right)={\log }_{0.5}\left(0.125\right)$

    • ${\log }_{0.5}$ cancels $0.5^{◻}$: $x={\log }_{0.5}\left(0.125\right)=3$

  • $0.5<1$ so $0.5^x$ is a decreasing function

    • That means $0.5^x>0.125$ for $x<3$

• E.g. $e^{2x}>7$

  • Start by solving as an equation $e^{2x}=7$

    • take $\ln$ of both sides: $\ln \left(e^{2x}\right)=\ln 7$

    • $\ln$ cancels $e^{◻}$: $2x=\ln 7$

  • $e>1$ so $e^{2x}$ is an increasing function

    • That means $e^{2x}>7$ for $2x>\ln 7$

  • Divide by 2 for the final solution

    • $x>\frac{\ln 7}{2}$

How can logarithms help reveal useful information about exponential expressions?

• The identity $b^x=c^{\left(x{\log }_{c}b\right)}$ allows you to convert between bases for an exponential function

  • This is covered in the Logarithmic Functions as Inverses of Exponential Functions study guide

• This is useful when you need to compare or combine exponential expressions with different bases

  • For example, $2^x$ can be rewritten as $e^{x\ln 2}$

  • This can be useful for solving equations that mix different exponential bases