Equations with power functions

Basic functions: Power functions

Equations with power functions

Finding real solutions of an equation with a power function of the form $f(x)=c\cdot x^p=y$ is always approached in the same way, namely ``isolate $x^p$ and then eliminate the $p$ -th power of $x^p$ by taking the $(1/p)$ -th power." To isolate the correct power, you sometimes have to convert square root notation in an equation into power notation You also have to keep an eye on whether you do not lose or introduce solutions in between.

For an equation involving a power function with a natural number $n$ unequal to $1$ as its exponent, the solution method boils down to using the $n$ -th root of the power. You usually need a calculator to compute numerical values.

The graph of the function $f(x)=x^n$ with $n$ a natural number greater than $1$ depends on whether the exponent $n$ is even or odd. You can see this in the interactive diagram below. In this diagram, also the horizontal line $y=c$ , for a certain number $c$ , can be manipulated. By playing with this it is easy to see how many real solutions the equation $x^n=c$ has.

even exponent (even $n$ )

GeoGebra

The equation has no solution if $c<0$ .
There are two solutions if $c>0$ , viz., $x=\pm\sqrt[n]{c}$ .
If $c=0$ , then the only solution is $x=0$ .

odd exponent (odd $n$ )

GeoGebra

The equation always has one solution.
If $c=0$ , then $x=0$ .
If $c\neq 0$ , then $x=\sqrt[n]{c}$ .

Find the exact real solution(s) of the following equation: $3\,x^2=27$

$x=-3\;\;\lor\;\;x=3$

$\begin{aligned} 3\,x^2&=27 &\phantom{abc}\blue{\text{the original equation}} \\[0.25cm]x^2&=9&\phantom{abc}\blue{\text{both sides divided by }3} \\[0.25cm] x&=\pm\sqrt[2]{9} &\phantom{abc}\blue{\text{taking the }2\text{nd root}}\\[0.25cm] x&=\pm\sqrt[2]{3^2}&\phantom{abc}\blue{\text{number under root symbol as power}}\\[0.25cm] x=-3\;\;\lor\;\;x&=3&\phantom{abc}\blue{\text{simplification}}\end{aligned}$

New example

The graph of the function $f(x)=x^p$ with a positive irriducible fraction $p=\tfrac{t}{n}$ as exponent depends on whether $n$ is even or odd, and for odd $n$ also on whether $t$ is even or odd. You can see this in the interactive diagram below. In this diagram, also the horizontal line $y=c$ , for a certain number $c$ , can be manipulated. By playing with this you can quickly see how many real solutions the equation $x^{\frac{t}{n}}=c$ has.

even denominator in the exponent (even $n$ )

GeoGebra

The equation has no solution if $c<0$ .
If $c>0$ , then one solution: $x=c^{\frac{n}{t}}=\sqrt[t]{c^n}$ .
If $c=0$ , then one solution: $x=0$ .

odd denominator in the exponent (odd $n$ )

GeoGebra

If $c=0$ , then $x=0$ .
If $c>0$ and $t$ odd, then $x=c^{\frac{n}{t}}=\sqrt[t]{c^n}$ .
If $c>0$ and $t$ are even, then two solutions: $x=c^{\frac{n}{t}}=\sqrt[t]{c^n}$ and $x=-c^{\frac{n}{t}}=-\sqrt[t]{c^n}$ .
If $c<0$ and $t$ odd, then $x=c^{\frac{n}{t}}=\sqrt[t]{c^n}$ .
If $c<0$ and $t$ are even, then there are no solutions.

Always check your answers!

A warning must be made for equations involving power functions with fractional exponents: If you do not check afterwards the solutions found in the original equation, you may have introduced apparent solutions. In reality, they are impossible, sometimes just because you created a number outside the domain of a power function.

A simple example: the equation $\sqrt{x}=-2$ has no real solution, but squaring left- and right-hand sides gives $x=(-2)^2=4$ . But this is not a solution of the original equation!

Solve the following equation exactly within the set of real numbers. $\sqrt[3]{x^2}=3$

$x=3\sqrt{3}\;\;\lor\;\; x=-3\sqrt{3}$

$\begin{aligned}\sqrt[3]{x^2} &= 3&\\[0.25cm] x^{\frac{2}{3}} &= 3& \phantom{abc}\blue{\text{power notation}}\\[0.25cm] x&=\pm(3)^{\frac{3}{2}}&\phantom{abc}\blue{\text{solution by exponentiation}}\\[0.25cm] x&=\pm\sqrt[2]{3^3}&\phantom{abc}\blue{\text{root noation}} \\[0.25cm]x=\sqrt{27}\;\;\lor\;\; x&=-\sqrt{27}&\phantom{abc}\blue{\text{simplification}} \\[0.25cm]x=3\sqrt{3}\;\;\lor\;\; x&=-3\sqrt{3} & \phantom{abc}\blue{\text{standard form}}\end{aligned}$ Plugging into the original equation shows that the solutions found are valid. The exponentiation in the solution method has not introduced an invalid solution.

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