Solve the following equations.
4x^3= 108
The equation can be modified:
4x^3=108
\dfrac{1}{4}\times\left(4x^3\right)= \dfrac{1}{4}\times 108
x^3=27
The equation can be solved as follows:
x=\sqrt[3]{27}=3
The solution is x=3.
2x^4=5x^4-48
The equation can be modified:
2x^4=5x^4-48
5x^4-2x^4=48
3x^4=48
\dfrac{1}{3} \times 3x^4=\dfrac{1}{3} \times 48
x^4=16
The equation can be solved as follows:
x=\sqrt[4]{16}=\pm2
The solutions are x=2 and x=-2.
2x^5= 64
The equation can be modified:
2x^5=64
\dfrac{1}{2}\times\left(2x^5\right)= \dfrac{1}{2}\times 64
x^5=32
The equation can be solved as follows:
x=\sqrt[5]{32}=2
The solution is x=2.
2x^2-8=4x^2
The equation can be modified:
2x^2-8=4x^2
2x^2-4x^2=8
-2x^2=8
x^2=-4
Since the square of every real number is positive, we can conclude that the equation has no solution in \mathbb{R}.
The equation has no solution in \mathbb{R}.
x^3-1=2x^3
The equation can be modified:
x^3-1=2x^3
2x^3-x^3=-1
x^3=-1
The equation can be solved as follows:
x=\sqrt[3]{-1}=-1
The solution is x=-1.
2x^2 = x^3
The equation can be modified:
2x^2=x^3
x^2\left(x-2\right)=0
Therefore either:
\begin{cases} x^2=0 \cr \text{or} \cr x-2=0 \end{cases}
We have:
\begin{cases} x=0 \cr \text{or} \cr x=2 \end{cases}
The solutions are x=0 and x=2.
2x^4=162
The equation can be modified:
2x^4=162
\dfrac{1}{2}\times\left(2x^4\right)= \dfrac{1}{2}\times 162
x^4=81
The equation can be solved as follows:
x=\sqrt[4]{81}=\pm3
The solutions are x=\pm3.