Find the Roots (Zeros) h(x)=81-3^(x+1)

$h (x) = 81 - 3^{x + 1}$

Set $81 - 3^{x + 1}$ equal to $0$ .

$81 - 3^{x + 1} = 0$

Solve for $x$ .

Subtract $81$ from both sides of the equation.

$- 3^{x + 1} = - 81$

Divide each term in $- 3^{x + 1} = - 81$ by $- 1$ and simplify.

Divide each term in $- 3^{x + 1} = - 81$ by $- 1$ .

$\frac{- 3^{x + 1}}{- 1} = \frac{- 81}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{3^{x + 1}}{1} = \frac{- 81}{- 1}$

Divide $3^{x + 1}$ by $1$ .

$3^{x + 1} = \frac{- 81}{- 1}$

Simplify the right side.

Divide $- 81$ by $- 1$ .

$3^{x + 1} = 81$

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

$3^{x + 1} = 3^{4}$

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

$x + 1 = 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$x = 4 - 1$

Subtract $1$ from $4$ .

$x = 3$