Solve for x 3^(5x)=27^(2x+1)

$3^{5 x} = 27^{2 x + 1}$

Step 1

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

$3^{5 x} = 3^{3 (2 x + 1)}$

Step 2

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

$5 x = 3 (2 x + 1)$

Step 3

Solve for $x$ .

Simplify $3 (2 x + 1)$ .

Apply the distributive property.

$5 x = 3 (2 x) + 3 \cdot 1$

Multiply.

Multiply $2$ by $3$ .

$5 x = 6 x + 3 \cdot 1$

Multiply $3$ by $1$ .

$5 x = 6 x + 3$

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

Subtract $6 x$ from both sides of the equation.

$5 x - 6 x = 3$

Subtract $6 x$ from $5 x$ .

$- x = 3$

Divide each term in $- x = 3$ by $- 1$ and simplify.

Divide each term in $- x = 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{3}{- 1}$

Simplify the right side.

Divide $3$ by $- 1$ .

$x = - 3$