Solve for x log base 4 of 5+ log base 4 of x = log base 4 of 7+ log base 4 of x-4

$\log_{4} (5) + \log_{4} (x) = \log_{4} (7) + \log_{4} (x - 4)$

Step 1

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

$\log_{4} (5 x) = \log_{4} (7) + \log_{4} (x - 4)$

Step 2

Simplify the right side.

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

$\log_{4} (5 x) = \log_{4} (7 (x - 4))$

Apply the distributive property.

$\log_{4} (5 x) = \log_{4} (7 x + 7 \cdot - 4)$

Multiply $7$ by $- 4$ .

$\log_{4} (5 x) = \log_{4} (7 x - 28)$

Step 3

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

$5 x = 7 x - 28$

Step 4

Solve for $x$ .

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

Subtract $7 x$ from both sides of the equation.

$5 x - 7 x = - 28$

Subtract $7 x$ from $5 x$ .

$- 2 x = - 28$

Divide each term in $- 2 x = - 28$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 28$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 28}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 28}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 28}{- 2}$

Simplify the right side.

Divide $- 28$ by $- 2$ .

$x = 14$