Evaluate log base 3 of 2x^2 = log base 3 of 2+2 log base 3 of x

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

Simplify the right side.

Simplify $\log_{3} (2) + 2 \log_{3} (x)$ .

Simplify $2 \log_{3} (x)$ by moving $2$ inside the logarithm.

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

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

$\log_{3} (2 x^{2}) = \log_{3} (2 x^{2})$

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

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

Solve for $x$ .

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

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

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

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

$0 = 0$

Since $0 = 0$ , the equation will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$