Evaluate log base 5 of x+8+ log base 5 of x+4=1

$\log_{5} (x + 8) + \log_{5} (x + 4) = 1$

Step 1

Simplify the left side.

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

$\log_{5} ((x + 8) (x + 4)) = 1$

Expand $(x + 8) (x + 4)$ using the FOIL Method.

Apply the distributive property.

$\log_{5} (x (x + 4) + 8 (x + 4)) = 1$

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 4 + 8 (x + 4)) = 1$

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 4 + 8 x + 8 \cdot 4) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{5} (x^{2} + x \cdot 4 + 8 x + 8 \cdot 4) = 1$

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

$\log_{5} (x^{2} + 4 \cdot x + 8 x + 8 \cdot 4) = 1$

Multiply $8$ by $4$ .

$\log_{5} (x^{2} + 4 x + 8 x + 32) = 1$

Add $4 x$ and $8 x$ .

$\log_{5} (x^{2} + 12 x + 32) = 1$

Step 2

Rewrite $\log_{5} (x^{2} + 12 x + 32) = 1$ 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 $b^{y} = x$ .

$5^{1} = x^{2} + 12 x + 32$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} + 12 x + 32 = 5^{1}$ .

$x^{2} + 12 x + 32 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} + 12 x + 32 - 5 = 0$

Subtract $5$ from $32$ .

$x^{2} + 12 x + 27 = 0$

Factor $x^{2} + 12 x + 27$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $27$ and whose sum is $12$ .

$3, 9$

Write the factored form using these integers.

$(x + 3) (x + 9) = 0$

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

$x + 3 = 0$

$x + 9 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $(x + 3) (x + 9) = 0$ true.

$x = - 3, - 9$

Step 4

Exclude the solutions that do not make $\log_{5} (x + 8) + \log_{5} (x + 4) = 1$ true.

$x = - 3$

Evaluate log base 5 of x+8+ log base 5 of x+4=1

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