### Functions: Power functions

### Equations of power functions

When it comes to power functions, you can also solve equations by means of factorization. For this end, you can use the following rule.

Higher Power Roots

If #a# is a non-negative real number, then there is exactly one non-negative real number #b#, in such a way that #b^n = a#. This number is written as #\sqrt[n]{a}#; it is called the #n#-**th power root** of #a#.

Solutions #b# of #b^n = a# | #a\gt 0# | #a=0# | #a\lt0# |

#n# even | #b=\pm\sqrt[n]{a}# | 0 | - |

#n# odd | #b=\sqrt[n]{a}# | 0 | #b=-\sqrt[n]{a}# |

#x=\sqrt[4]{{{1}\over{3}}}#

We will be working towards an equation without roots.

\[

\begin{array}{rclcl}

6x^{{{7}\over{2}}}&=&\dfrac{2}{\sqrt{x}}&\phantom{xxx}&\color{blue}{\text{the original equation}}\\

6x^{{{7}\over{2}}}\cdot x^{\frac{1}{2}} &= &2&\phantom{xxx}&\color{blue}{\text{multiplied by }\sqrt{x}=x^{\frac{1}{2}}}\\

6x^{4}&=&2&\phantom{xxx}&\color{blue}{\text{simplified with the help of rules of calculation for powers }}\\

x^{4}&=&{{1}\over{3}}&\phantom{xxx}&\color{blue}{\text{divided by }6\tiny.}\\

\end{array}

\]

Because #4# is even, there are two solutions, being: #x=\sqrt[4]{{{1}\over{3}}}\lor x=-\sqrt[4]{{{1}\over{3}}}#. These two values hence are solution to the equation #x^{4}={{1}\over{3}}#, but not necessarily to the original equation #6x^{{{7}\over{2}}}=\dfrac{2}{\sqrt{x}}#. After all, the original equation is only defined if #x\gt0#. The negative solution therefore #x=-\sqrt[4]{{{1}\over{3}}}# is not valid, and the only solution is #x=\sqrt[4]{{{1}\over{3}}}#.

We will be working towards an equation without roots.

\[

\begin{array}{rclcl}

6x^{{{7}\over{2}}}&=&\dfrac{2}{\sqrt{x}}&\phantom{xxx}&\color{blue}{\text{the original equation}}\\

6x^{{{7}\over{2}}}\cdot x^{\frac{1}{2}} &= &2&\phantom{xxx}&\color{blue}{\text{multiplied by }\sqrt{x}=x^{\frac{1}{2}}}\\

6x^{4}&=&2&\phantom{xxx}&\color{blue}{\text{simplified with the help of rules of calculation for powers }}\\

x^{4}&=&{{1}\over{3}}&\phantom{xxx}&\color{blue}{\text{divided by }6\tiny.}\\

\end{array}

\]

Because #4# is even, there are two solutions, being: #x=\sqrt[4]{{{1}\over{3}}}\lor x=-\sqrt[4]{{{1}\over{3}}}#. These two values hence are solution to the equation #x^{4}={{1}\over{3}}#, but not necessarily to the original equation #6x^{{{7}\over{2}}}=\dfrac{2}{\sqrt{x}}#. After all, the original equation is only defined if #x\gt0#. The negative solution therefore #x=-\sqrt[4]{{{1}\over{3}}}# is not valid, and the only solution is #x=\sqrt[4]{{{1}\over{3}}}#.

Some function, that include a power function, with unknown #x# can be reduced to simpler equations with unknown #y# by substitutions as #y=f(x)# for an appropriate (power) function #f#.

Solve the equation #x^{{{2}\over{5}}}-8\cdot x^{{{1}\over{5}}}+15=0# with unknown #x#, using the factorization \[x^{{{2}\over{5}}}-8\cdot x^{{{1}\over{5}}}+15=(x^{1/5}-5)(x^{1/5}-3)\tiny.\]Provide an answer of the form #x=a\vee x=b# if there are two solutions, of the form #x=a# if there is a single solution, and write #none# if there are no solutions.

#x=3125\lor x=243#

This can be seen as follows.\[\begin{array}{rclcl}x^{{{2}\over{5}}}-8\cdot x^{{{1}\over{5}}}+15&=&0&\phantom{x}&\color{blue}{\text{the original equation}}\\

y^2-8\cdot y+15 &=& 0&\phantom{x}&\color{blue}{\text{substituted }y=x^{1/5}}\\

\left(y-5\right)\left(y-3\right)&=&0&\phantom{x}&\color{blue}{\text{factorization}}\\

y-5=0&\lor&y-3=0&\phantom{x}&\color{blue}{\text{split according to factors}}\\

y=5&\lor& y=3&\phantom{x}&\color{blue}{\text{constant terms to the right hand side}}\\{}

x=5^5&\lor& x=3^{5} &\phantom{x}&\color{blue}{\text{substitute back } y^5=x}\\

x=3125&\lor& x=243 &\phantom{x}&\color{blue}{\text{exponentiation to } 5}\end{array}\]

This can be seen as follows.\[\begin{array}{rclcl}x^{{{2}\over{5}}}-8\cdot x^{{{1}\over{5}}}+15&=&0&\phantom{x}&\color{blue}{\text{the original equation}}\\

y^2-8\cdot y+15 &=& 0&\phantom{x}&\color{blue}{\text{substituted }y=x^{1/5}}\\

\left(y-5\right)\left(y-3\right)&=&0&\phantom{x}&\color{blue}{\text{factorization}}\\

y-5=0&\lor&y-3=0&\phantom{x}&\color{blue}{\text{split according to factors}}\\

y=5&\lor& y=3&\phantom{x}&\color{blue}{\text{constant terms to the right hand side}}\\{}

x=5^5&\lor& x=3^{5} &\phantom{x}&\color{blue}{\text{substitute back } y^5=x}\\

x=3125&\lor& x=243 &\phantom{x}&\color{blue}{\text{exponentiation to } 5}\end{array}\]

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