Solve for x log of x+4 = log of x+ log of 4

$\log (x + 4) = \log (x) + \log (4)$

Step 1

Simplify the right side.

Simplify $\log (x) + \log (4)$ .

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

$\log (x + 4) = \log (x \cdot 4)$

Move $4$ to the left of $x$ .

$\log (x + 4) = \log (4 x)$

Step 2

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

$x + 4 = 4 x$

Step 3

Solve for $x$ .

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

Subtract $4 x$ from both sides of the equation.

$x + 4 - 4 x = 0$

Subtract $4 x$ from $x$ .

$- 3 x + 4 = 0$

Subtract $4$ from both sides of the equation.

$- 3 x = - 4$

Divide each term in $- 3 x = - 4$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 4$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 4}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 4}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 3}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{4}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{4}{3}$

Decimal Form:

$x = 1.\overline{3}$

Mixed Number Form:

$x = 1 \frac{1}{3}$