Solve for x 9^(2x+1)=81(27^x)

$9^{2 x + 1} = 81 (27^{x})$

Rewrite $81$ as $3^{4}$ .

$9^{2 x + 1} = 3^{4} \cdot 27^{x}$

Rewrite $27$ as $3^{3}$ .

$9^{2 x + 1} = 3^{4} \cdot {(3^{3})}^{x}$

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

$9^{2 x + 1} = 3^{4} \cdot 3^{3 x}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$9^{2 x + 1} = 3^{4 + 3 x}$

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

$3^{2 (2 x + 1)} = 3^{4 + 3 x}$

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

$2 (2 x + 1) = 4 + 3 x$

Solve for $x$ .

Simplify $2 (2 x + 1)$ .

Rewrite.

$0 + 0 + 2 (2 x + 1) = 4 + 3 x$

Simplify by adding zeros.

$2 (2 x + 1) = 4 + 3 x$

Apply the distributive property.

$2 (2 x) + 2 \cdot 1 = 4 + 3 x$

Multiply.

Multiply $2$ by $2$ .

$4 x + 2 \cdot 1 = 4 + 3 x$

Multiply $2$ by $1$ .

$4 x + 2 = 4 + 3 x$

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

Subtract $3 x$ from both sides of the equation.

$4 x + 2 - 3 x = 4$

Subtract $3 x$ from $4 x$ .

$x + 2 = 4$

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

Subtract $2$ from both sides of the equation.

$x = 4 - 2$

Subtract $2$ from $4$ .

$x = 2$