Solve for x log of x+ log of x+3=2 log of x+1

$\log (x) + \log (x + 3) = 2 \log (x + 1)$

Simplify the left side.

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

$\log (x (x + 3)) = 2 \log (x + 1)$

Apply the distributive property.

$\log (x \cdot x + x \cdot 3) = 2 \log (x + 1)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot 3) = 2 \log (x + 1)$

Move $3$ to the left of $x$ .

$\log (x^{2} + 3 x) = 2 \log (x + 1)$

Simplify $2 \log (x + 1)$ by moving $2$ inside the logarithm.

$\log (x^{2} + 3 x) = \log ({(x + 1)}^{2})$

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

$x^{2} + 3 x = {(x + 1)}^{2}$

Solve for $x$ .

Simplify ${(x + 1)}^{2}$ .

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$x^{2} + 3 x = (x + 1) (x + 1)$

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

Apply the distributive property.

$x^{2} + 3 x = x (x + 1) + 1 (x + 1)$

Apply the distributive property.

$x^{2} + 3 x = x \cdot x + x \cdot 1 + 1 (x + 1)$

Apply the distributive property.

$x^{2} + 3 x = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 3 x = x^{2} + x \cdot 1 + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$x^{2} + 3 x = x^{2} + x + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$x^{2} + 3 x = x^{2} + x + x + 1 \cdot 1$

Multiply $1$ by $1$ .

$x^{2} + 3 x = x^{2} + x + x + 1$

Add $x$ and $x$ .

$x^{2} + 3 x = x^{2} + 2 x + 1$

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

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

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

Subtract $2 x$ from both sides of the equation.

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

Combine the opposite terms in $x^{2} + 3 x - x^{2} - 2 x$ .

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

$3 x + 0 - 2 x = 1$

Add $3 x$ and $0$ .

$3 x - 2 x = 1$

Subtract $2 x$ from $3 x$ .

$x = 1$

Solve for x log of x+ log of x+3=2 log of x+1

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