Solve for x log base 6 of x+2+ log base 6 of x+3=1

$\log_{6} (x + 2) + \log_{6} (x + 3) = 1$

Step 1

Simplify the left side.

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

$\log_{6} ((x + 2) (x + 3)) = 1$

Expand $(x + 2) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$\log_{6} (x (x + 3) + 2 (x + 3)) = 1$

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot 3 + 2 (x + 3)) = 1$

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot 3 + 2 x + 2 \cdot 3) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{6} (x^{2} + x \cdot 3 + 2 x + 2 \cdot 3) = 1$

Move $3$ to the left of $x$ .

$\log_{6} (x^{2} + 3 \cdot x + 2 x + 2 \cdot 3) = 1$

Multiply $2$ by $3$ .

$\log_{6} (x^{2} + 3 x + 2 x + 6) = 1$

Add $3 x$ and $2 x$ .

$\log_{6} (x^{2} + 5 x + 6) = 1$

Step 2

Rewrite $\log_{6} (x^{2} + 5 x + 6) = 1$ 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 $b^{y} = x$ .

$6^{1} = x^{2} + 5 x + 6$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} + 5 x + 6 = 6^{1}$ .

$x^{2} + 5 x + 6 = 6$

Subtract $6$ from both sides of the equation.

$x^{2} + 5 x + 6 - 6 = 0$

Combine the opposite terms in $x^{2} + 5 x + 6 - 6$ .

Subtract $6$ from $6$ .

$x^{2} + 5 x + 0 = 0$

Add $x^{2} + 5 x$ and $0$ .

$x^{2} + 5 x = 0$

Factor $x$ out of $x^{2} + 5 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x + 5 x = 0$

Factor $x$ out of $5 x$ .

$x \cdot x + x \cdot 5 = 0$

Factor $x$ out of $x \cdot x + x \cdot 5$ .

$x (x + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $x (x + 5) = 0$ true.

$x = 0, - 5$

Step 4

Exclude the solutions that do not make $\log_{6} (x + 2) + \log_{6} (x + 3) = 1$ true.

$x = 0$

Solve for x log base 6 of x+2+ log base 6 of x+3=1

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