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

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

Step 1

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

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

Step 2

Simplify the right side.

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

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

Apply the distributive property.

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

Multiply $5$ by $- 2$ .

$\log_{2} (3 x) = \log_{2} (5 x - 10)$

Step 3

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

$3 x = 5 x - 10$

Step 4

Solve for $x$ .

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

Subtract $5 x$ from both sides of the equation.

$3 x - 5 x = - 10$

Subtract $5 x$ from $3 x$ .

$- 2 x = - 10$

Divide each term in $- 2 x = - 10$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 10$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 10}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 10}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 10}{- 2}$

Simplify the right side.

Divide $- 10$ by $- 2$ .

$x = 5$