Solve for x log base 5 of x+7=1- log base 5 of x+3

$\log_{5} (x + 7) = 1 - \log_{5} (x + 3)$

Move all the terms containing a logarithm to the left side of the equation.

$\log_{5} (x + 7) + \log_{5} (x + 3) = 1$

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

$\log_{5} ((x + 7) (x + 3)) = 1$

Expand $(x + 7) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$\log_{5} (x (x + 3) + 7 (x + 3)) = 1$

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 3 + 7 (x + 3)) = 1$

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot 3 + 7 x + 7 \cdot 3) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{5} (x^{2} + x \cdot 3 + 7 x + 7 \cdot 3) = 1$

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

$\log_{5} (x^{2} + 3 \cdot x + 7 x + 7 \cdot 3) = 1$

Multiply $7$ by $3$ .

$\log_{5} (x^{2} + 3 x + 7 x + 21) = 1$

Add $3 x$ and $7 x$ .

$\log_{5} (x^{2} + 10 x + 21) = 1$

Rewrite $\log_{5} (x^{2} + 10 x + 21) = 1$ 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 $b^{y} = x$ .

$5^{1} = x^{2} + 10 x + 21$

Solve for $x$ .

Rewrite the equation as $x^{2} + 10 x + 21 = 5^{1}$ .

$x^{2} + 10 x + 21 = 5$

Subtract $5$ from both sides of the equation.

$x^{2} + 10 x + 21 - 5 = 0$

Subtract $5$ from $21$ .

$x^{2} + 10 x + 16 = 0$

Factor $x^{2} + 10 x + 16$ 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 $16$ and whose sum is $10$ .

$2, 8$

Write the factored form using these integers.

$(x + 2) (x + 8) = 0$

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

$x + 2 = 0$

$x + 8 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(x + 2) (x + 8) = 0$ true.

$x = - 2, - 8$

Exclude the solutions that do not make $\log_{5} (x + 7) = 1 - \log_{5} (x + 3)$ true.

$x = - 2$

Solve for x log base 5 of x+7=1- log base 5 of x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.