Solve for x 3^(x^2)=81^x

$3^{x^{2}} = 81^{x}$

Step 1

Create equivalent expressions in the equation that all have equal bases.

$3^{x^{2}} = 3^{4 x}$

Step 2

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

$x^{2} = 4 x$

Step 3

Solve for $x$ .

Subtract $4 x$ from both sides of the equation.

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

Factor $x$ out of $x^{2} - 4 x$ .

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

$x \cdot x - 4 x = 0$

Factor $x$ out of $- 4 x$ .

$x \cdot x + x \cdot - 4 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

$x (x - 4) = 0$

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

$x = 0$

$x - 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $x (x - 4) = 0$ true.

$x = 0, 4$