Solve for x natural log of -3x-1- natural log of 7=2

$\ln (- 3 x - 1) - \ln (7) = 2$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{- 3 x - 1}{7}) = 2$

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{- 3 x - 1}{7})} = e^{2}$

Rewrite $\ln (\frac{- 3 x - 1}{7}) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{2} = \frac{- 3 x - 1}{7}$

Solve for $x$ .

Rewrite the equation as $\frac{- 3 x - 1}{7} = e^{2}$ .

$\frac{- 3 x - 1}{7} = e^{2}$

Multiply both sides by $7$ .

$\frac{- 3 x - 1}{7} \cdot 7 = e^{2} \cdot 7$

Simplify.

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{- 3 x - 1}{7} \cdot 7 = e^{2} \cdot 7$

Rewrite the expression.

$- 3 x - 1 = e^{2} \cdot 7$

Simplify the right side.

Move $7$ to the left of $e^{2}$ .

$- 3 x - 1 = 7 e^{2}$

Solve for $x$ .

Add $1$ to both sides of the equation.

$- 3 x = 7 e^{2} + 1$

Divide each term in $- 3 x = 7 e^{2} + 1$ by $- 3$ and simplify.

Divide each term in $- 3 x = 7 e^{2} + 1$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{7 e^{2}}{- 3} + \frac{1}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{7 e^{2}}{- 3} + \frac{1}{- 3}$

Divide $x$ by $1$ .

$x = \frac{7 e^{2}}{- 3} + \frac{1}{- 3}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x = - \frac{7 e^{2}}{3} + \frac{1}{- 3}$

Move the negative in front of the fraction.

$x = - \frac{7 e^{2}}{3} - \frac{1}{3}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{7 e^{2}}{3} - \frac{1}{3}$

Decimal Form:

$x = - 17.57446423 \dots$

Solve for x natural log of -3x-1- natural log of 7=2

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