Functions: Power functions
Equations with power functions
In quadratic equations we have seen how to solve an equation #x^2=c#. With the same procedure, we will use higher degree roots to solve an equation #x^n=c#.
The solutions to the equation #x^\orange{n}=\blue{c}# are dependent on the values of #\orange n# and #\blue c#.
#\blue{c} \gt 0#  #\blue{c}=0#  #\blue{c} \lt 0#  
#\orange{n}# is even 
Two solutions: #x=\sqrt[\orange{n}]{\blue{c}} \lor x=\sqrt[\orange{n}]{\blue{c}}# 
One solution: #x=0# 
No solutions

#\orange{n}# is odd 
One solution: #x=\sqrt[\orange{n}]{\blue{c}}# 
One solution: #x=0# 
One solution: #x=\sqrt[\orange{n}]{\blue{c}}# 
In the examples we see that you can reduce many equations to the form #x^\orange{n}=\blue{c}# and then solve them.
#x=\frac{1}{2}\sqrt[6]{48} \lor x=\frac{1}{2}\sqrt[6]{48}#
#\begin{array}{rcl}4\, x^{6}+6&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
4\, x^{6}&=&3 \\
&&\phantom{xxx}\blue{\text{both sides minus }6} \\
x^{6} &=& {{3}\over{4}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }4} \\
x=\sqrt[6]{{{3}\over{4}}} &\lor& x=\sqrt[6]{{{3}\over{4}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }6 \text{th root}}\\
x=\frac{1}{2}\sqrt[6]{48} &\lor& x=\frac{1}{2}\sqrt[6]{48}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
#\begin{array}{rcl}4\, x^{6}+6&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
4\, x^{6}&=&3 \\
&&\phantom{xxx}\blue{\text{both sides minus }6} \\
x^{6} &=& {{3}\over{4}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }4} \\
x=\sqrt[6]{{{3}\over{4}}} &\lor& x=\sqrt[6]{{{3}\over{4}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }6 \text{th root}}\\
x=\frac{1}{2}\sqrt[6]{48} &\lor& x=\frac{1}{2}\sqrt[6]{48}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
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