Solve for x natural log of x-1+ natural log of x+2=1

$\ln (x - 1) + \ln (x + 2) = 1$

Step 1

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln ((x - 1) (x + 2)) = 1$

Expand $(x - 1) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$\ln (x (x + 2) - 1 (x + 2)) = 1$

Apply the distributive property.

$\ln (x \cdot x + x \cdot 2 - 1 (x + 2)) = 1$

Apply the distributive property.

$\ln (x \cdot x + x \cdot 2 - 1 x - 1 \cdot 2) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 2 - 1 x - 1 \cdot 2) = 1$

Move $2$ to the left of $x$ .

$\ln (x^{2} + 2 \cdot x - 1 x - 1 \cdot 2) = 1$

Rewrite $- 1 x$ as $- x$ .

$\ln (x^{2} + 2 x - x - 1 \cdot 2) = 1$

Multiply $- 1$ by $2$ .

$\ln (x^{2} + 2 x - x - 2) = 1$

Subtract $x$ from $2 x$ .

$\ln (x^{2} + x - 2) = 1$

Step 2

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

$e^{\ln (x^{2} + x - 2)} = e^{1}$

Step 3

Rewrite $\ln (x^{2} + x - 2) = 1$ 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 $b^{y} = x$ .

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

Step 4

Solve for $x$ .

Rewrite the equation as $x^{2} + x - 2 = e^{1}$ .

$x^{2} + x - 2 = e$

Subtract $e$ from both sides of the equation.

$x^{2} + x - 2 - e = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = - 2 - e$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot (- 2 - e))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 2 - e)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (- 2 - e)}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 2 - 4 (- e)}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$x = \frac{- 1 \pm \sqrt{1 + 8 - 4 (- e)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 8 + 4 e}}{2 \cdot 1}$

Add $1$ and $8$ .

$x = \frac{- 1 \pm \sqrt{9 + 4 e}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{9 + 4 e}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{9 + 4 e}}{2}, - \frac{1 + \sqrt{9 + 4 e}}{2}$

Step 5

Exclude the solutions that do not make $\ln (x - 1) + \ln (x + 2) = 1$ true.

$x = - \frac{1 - \sqrt{9 + 4 e}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{9 + 4 e}}{2}$

Decimal Form:

$x = 1.72896429 \dots$

Solve for x natural log of x-1+ natural log of x+2=1

Related Questions

Support LCMGCF.com by disabling your adblocker.