Solve for x log base 3 of 2x- log base 3 of x+5=0

$\log_{3} (2 x) - \log_{3} (x + 5) = 0$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{3} (\frac{2 x}{x + 5}) = 0$

Rewrite $\log_{3} (\frac{2 x}{x + 5}) = 0$ 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 $b^{y} = x$ .

$3^{0} = \frac{2 x}{x + 5}$

Cross multiply to remove the fraction.

$2 x = 3^{0} (x + 5)$

Simplify $3^{0} (x + 5)$ .

Anything raised to $0$ is $1$ .

$2 x = 1 (x + 5)$

Multiply $x + 5$ by $1$ .

$2 x = x + 5$

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

Subtract $x$ from both sides of the equation.

$2 x - x = 5$

Subtract $x$ from $2 x$ .

$x = 5$

Solve for x log base 3 of 2x- log base 3 of x+5=0

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