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

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

Step 1

Simplify the left side.

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

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

Simplify by multiplying through.

Apply the distributive property.

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

Simplify the expression.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

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

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

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} - x = 2$

Step 3

Solve for $x$ .

Subtract $2$ from both sides of the equation.

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

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$(x - 2) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

$x + 1 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Step 4

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

$x = 2$

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

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