Solve for x (4^x)^(x-3)=256

${(4^{x})}^{x - 3} = 256$

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

$4^{x (x - 3)} = 256$

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

$4^{x (x - 3)} = 4^{4}$

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

$x (x - 3) = 4$

Solve for $x$ .

Simplify $x (x - 3)$ .

Apply the distributive property.

$x \cdot x + x \cdot - 3 = 4$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 = 4$

Move $- 3$ to the left of $x$ .

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

Subtract $4$ from both sides of the equation.

$x^{2} - 3 x - 4 = 0$

Factor $x^{2} - 3 x - 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

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.

$x = 4$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve for x (4^x)^(x-3)=256

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