Solve for x log of x+ log of x+1 = log of x^2+2

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

Simplify the left side.

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

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

Apply the distributive property.

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

Simplify the expression.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

$\log (x^{2} + x) = \log (x^{2} + 2)$

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

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

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

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

Combine the opposite terms in $x^{2} + x - x^{2}$ .

Subtract $x^{2}$ from $x^{2}$ .

$x + 0 = 2$

Add $x$ and $0$ .

$x = 2$