Solve for x 3^(2x+1)=(1/3)^x

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

Step 1

Apply the product rule to $\frac{1}{3}$ .

$3^{2 x + 1} = \frac{1^{x}}{3^{x}}$

Step 2

One to any power is one.

$3^{2 x + 1} = \frac{1}{3^{x}}$

Step 3

Move $3^{x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$3^{2 x + 1} = 3^{- x}$

Step 4

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

$2 x + 1 = - x$

Step 5

Solve for $x$ .

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

Add $x$ to both sides of the equation.

$2 x + 1 + x = 0$

Add $2 x$ and $x$ .

$3 x + 1 = 0$

Subtract $1$ from both sides of the equation.

$3 x = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.\overline{3}$