Solve for x 3-4 natural log of 8x+1=12

$3 - 4 \ln (8 x + 1) = 12$

Move all terms not containing $\ln (8 x + 1)$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$- 4 \ln (8 x + 1) = 12 - 3$

Subtract $3$ from $12$ .

$- 4 \ln (8 x + 1) = 9$

Divide each term in $- 4 \ln (8 x + 1) = 9$ by $- 4$ and simplify.

Divide each term in $- 4 \ln (8 x + 1) = 9$ by $- 4$ .

$\frac{- 4 \ln (8 x + 1)}{- 4} = \frac{9}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 \ln (8 x + 1)}{- 4} = \frac{9}{- 4}$

Divide $\ln (8 x + 1)$ by $1$ .

$\ln (8 x + 1) = \frac{9}{- 4}$

Simplify the right side.

Move the negative in front of the fraction.

$\ln (8 x + 1) = - \frac{9}{4}$

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

$e^{\ln (8 x + 1)} = e^{- \frac{9}{4}}$

Rewrite $\ln (8 x + 1) = - \frac{9}{4}$ 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 $b^{y} = x$ .

$e^{- \frac{9}{4}} = 8 x + 1$

Solve for $x$ .

Rewrite the equation as $8 x + 1 = e^{- \frac{9}{4}}$ .

$8 x + 1 = e^{- \frac{9}{4}}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$8 x + 1 = \frac{1}{e^{\frac{9}{4}}}$

Subtract $1$ from both sides of the equation.

$8 x = \frac{1}{e^{\frac{9}{4}}} - 1$

Divide each term in $8 x = \frac{1}{e^{\frac{9}{4}}} - 1$ by $8$ and simplify.

Divide each term in $8 x = \frac{1}{e^{\frac{9}{4}}} - 1$ by $8$ .

$\frac{8 x}{8} = \frac{\frac{1}{e^{\frac{9}{4}}}}{8} + \frac{- 1}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{\frac{1}{e^{\frac{9}{4}}}}{8} + \frac{- 1}{8}$

Divide $x$ by $1$ .

$x = \frac{\frac{1}{e^{\frac{9}{4}}}}{8} + \frac{- 1}{8}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{1}{e^{\frac{9}{4}}} \cdot \frac{1}{8} + \frac{- 1}{8}$

Combine.

$x = \frac{1 \cdot 1}{e^{\frac{9}{4}} \cdot 8} + \frac{- 1}{8}$

Multiply $1$ by $1$ .

$x = \frac{1}{e^{\frac{9}{4}} \cdot 8} + \frac{- 1}{8}$

Move $8$ to the left of $e^{\frac{9}{4}}$ .

$x = \frac{1}{8 \cdot e^{\frac{9}{4}}} + \frac{- 1}{8}$

Move the negative in front of the fraction.

$x = \frac{1}{8 e^{\frac{9}{4}}} - \frac{1}{8}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{8 e^{\frac{9}{4}}} - \frac{1}{8}$

Decimal Form:

$x = - 0.11182509 \dots$

Solve for x 3-4 natural log of 8x+1=12

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