Solve for x log of 4+ log of x+2=1

$\log (4) + \log (x + 2) = 1$

Step 1

Simplify the left side.

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

$\log (4 (x + 2)) = 1$

Apply the distributive property.

$\log (4 x + 4 \cdot 2) = 1$

Multiply $4$ by $2$ .

$\log (4 x + 8) = 1$

Step 2

Rewrite $\log (4 x + 8) = 1$ 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 $b^{y} = x$ .

$10^{1} = 4 x + 8$

Step 3

Solve for $x$ .

Rewrite the equation as $4 x + 8 = 10^{1}$ .

$4 x + 8 = 10$

Move all terms not containing $x$ to the right side of the equation.

Subtract $8$ from both sides of the equation.

$4 x = 10 - 8$

Subtract $8$ from $10$ .

$4 x = 2$

Divide each term in $4 x = 2$ by $4$ and simplify.

Divide each term in $4 x = 2$ by $4$ .

$\frac{4 x}{4} = \frac{2}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{2}{4}$

Divide $x$ by $1$ .

$x = \frac{2}{4}$

Simplify the right side.

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$x = \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot 1}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot 1}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{1}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}$

Decimal Form:

$x = 0.5$

Solve for x log of 4+ log of x+2=1

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