Solve for x 81^(5x)=27^(x+1)

$81^{5 x} = 27^{x + 1}$

Step 1

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

$3^{4 (5 x)} = 3^{3 (x + 1)}$

Step 2

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

$4 (5 x) = 3 (x + 1)$

Step 3

Solve for $x$ .

Multiply $5$ by $4$ .

$20 x = 3 (x + 1)$

Simplify $3 (x + 1)$ .

Apply the distributive property.

$20 x = 3 x + 3 \cdot 1$

Multiply $3$ by $1$ .

$20 x = 3 x + 3$

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

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

$20 x - 3 x = 3$

Subtract $3 x$ from $20 x$ .

$17 x = 3$

Divide each term in $17 x = 3$ by $17$ and simplify.

Divide each term in $17 x = 3$ by $17$ .

$\frac{17 x}{17} = \frac{3}{17}$

Simplify the left side.

Cancel the common factor of $17$ .

Cancel the common factor.

$\frac{17 x}{17} = \frac{3}{17}$

Divide $x$ by $1$ .

$x = \frac{3}{17}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3}{17}$

Decimal Form:

$x = 0.17647058 \dots$