Solve for r 3^(r+1)=1

$3^{r + 1} = 1$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3^{r + 1}) = \ln (1)$

Step 2

Expand $\ln (3^{r + 1})$ by moving $r + 1$ outside the logarithm.

$(r + 1) \ln (3) = \ln (1)$

Step 3

Simplify the left side.

Simplify $(r + 1) \ln (3)$ .

Apply the distributive property.

$r \ln (3) + 1 \ln (3) = \ln (1)$

Multiply $\ln (3)$ by $1$ .

$r \ln (3) + \ln (3) = \ln (1)$

Step 4

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$r \ln (3) + \ln (3) = 0$

Step 5

Subtract $\ln (3)$ from both sides of the equation.

$r \ln (3) = - \ln (3)$

Step 6

Divide each term in $r \ln (3) = - \ln (3)$ by $\ln (3)$ and simplify.

Divide each term in $r \ln (3) = - \ln (3)$ by $\ln (3)$ .

$\frac{r \ln (3)}{\ln (3)} = \frac{- \ln (3)}{\ln (3)}$

Simplify the left side.

Cancel the common factor of $\ln (3)$ .

Cancel the common factor.

$\frac{r \ln (3)}{\ln (3)} = \frac{- \ln (3)}{\ln (3)}$

Divide $r$ by $1$ .

$r = \frac{- \ln (3)}{\ln (3)}$

Simplify the right side.

Cancel the common factor of $\ln (3)$ .

Cancel the common factor.

$r = \frac{- \ln (3)}{\ln (3)}$

Divide $- 1$ by $1$ .

$r = - 1$