Solve for x (x^2)^x=x^(4x-x^2)

${(x^{2})}^{x} = x^{4 x - x^{2}}$

Step 1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2 x} = x^{4 x - x^{2}}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 x = 4 x - x^{2}$

Step 3

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$4 x - x^{2} = 2 x$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$4 x - x^{2} - 2 x = 0$

Subtract $2 x$ from $4 x$ .

$- x^{2} + 2 x = 0$

Factor $- x$ out of $- x^{2} + 2 x$ .

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 2 x = 0$

Factor $- x$ out of $2 x$ .

$- x \cdot x - x \cdot - 2 = 0$

Factor $- x$ out of $- x (x) - x (- 2)$ .

$- x (x - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $- x (x - 2) = 0$ true.

$x = 0, 2$