Solving Equations
1.6 Other Types of Equations
Rational Exponents
Rational exponents connect powers and roots. Once this connection is clear, many equations become routine.
Definition: Rational Exponent
For $a>0$ and integers $m,n$ with $n>0$,
$$a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m.$$
Example: Evaluate rational exponents
Evaluate.
- $8^{\frac{2}{3}}$
- $16^{\frac{3}{4}}$
Show Solution
- $8^{\frac{2}{3}}=(\sqrt[3]{8})^2=2^2=4.$
- $16^{\frac{3}{4}}=(\sqrt[4]{16})^3=2^3=8.$
MyOpenMath: Practice: rational exponents (placeholder)
Solving Equations with Rational Exponents
If the variable has a fractional exponent, apply the inverse power to both sides.
Fact: Inverse-power strategy
If
$$x^{\frac{p}{q}}=k,$$
raise both sides to the power $\frac{q}{p}$ (when defined) to isolate $x$.
Example: Solve a rational-exponent equation
Solve:
$$x^{\frac{3}{5}}=8.$$
Show Solution
Raise both sides to the reciprocal power $\frac{5}{3}$:
$$x=8^{\frac{5}{3}}=(\sqrt[3]{8})^5=2^5=32.$$
Final answer:
$$x=32. $$
Example: Solve by factoring after substitution
Solve:
$$3x^{\frac{3}{5}}=x^{\frac{1}{5}}.$$
Show Solution
Move all terms to one side:
$$3x^{\frac{3}{5}}-x^{\frac{1}{5}}=0.$$
Factor out $x^{\frac{1}{5}}$:
$$x^{\frac{1}{5}}\left(3x^{\frac{2}{5}}-1\right)=0.$$
So either
$$x^{\frac{1}{5}}=0\Rightarrow x=0,$$
or
$$3x^{\frac{2}{5}}-1=0\Rightarrow x^{\frac{2}{5}}=\frac13.$$
Raise both sides to $\frac52$:
$$x=\left(\frac13\right)^{\frac52}=\frac{1}{9\sqrt{3}}. $$
Final solution set:
$$\left{0,\left(\frac13\right)^{\frac52}\right}. $$
MyOpenMath: Practice: solving rational-exponent equations (placeholder)
Radical Equations and Extraneous Solutions
Squaring both sides can create fake solutions, so checking is mandatory.
Warning: Extraneous-solution warning
Whenever you square both sides of an equation, check every candidate in the original equation.
Example: Solve a radical equation
Solve:
$$\sqrt{x+2}=x.$$
Show Solution
Square both sides:
$$x+2=x^2\Rightarrow x^2-x-2=0.$$
Factor:
$$(x-2)(x+1)=0\Rightarrow x=2\text{ or }x=-1.$$
Check in original equation:
- $x=2$: $\sqrt{4}=2$, valid.
- $x=-1$: $\sqrt{1}=-1$, false.
So the only solution is
$$x=2. $$
MyOpenMath: Practice: radical equations and checks (placeholder)
Absolute Value Equations
Absolute value equations usually split into two linear equations.
Definition: Absolute Value Equation Pattern
If $b\ge 0$, then
$$|u|=b\iff u=b\text{ or }u=-b.$$
Example: Solve an absolute value equation
Solve:
$$|2x-3|=7.$$
Show Solution
Split into two cases:
- $2x-3=7\Rightarrow 2x=10\Rightarrow x=5$
- $2x-3=-7\Rightarrow 2x=-4\Rightarrow x=-2$
Final solution set:
$${-2,5}. $$