Solve for x log base 9 of x+6+ log base 9 of x+4=1/2

$\log_{9} (x + 6) + \log_{9} (x + 4) = \frac{1}{2}$

Simplify the left side.

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

$\log_{9} ((x + 6) (x + 4)) = \frac{1}{2}$

Expand $(x + 6) (x + 4)$ using the FOIL Method.

Apply the distributive property.

$\log_{9} (x (x + 4) + 6 (x + 4)) = \frac{1}{2}$

Apply the distributive property.

$\log_{9} (x \cdot x + x \cdot 4 + 6 (x + 4)) = \frac{1}{2}$

Apply the distributive property.

$\log_{9} (x \cdot x + x \cdot 4 + 6 x + 6 \cdot 4) = \frac{1}{2}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{9} (x^{2} + x \cdot 4 + 6 x + 6 \cdot 4) = \frac{1}{2}$

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

$\log_{9} (x^{2} + 4 \cdot x + 6 x + 6 \cdot 4) = \frac{1}{2}$

Multiply $6$ by $4$ .

$\log_{9} (x^{2} + 4 x + 6 x + 24) = \frac{1}{2}$

Add $4 x$ and $6 x$ .

$\log_{9} (x^{2} + 10 x + 24) = \frac{1}{2}$

Rewrite $\log_{9} (x^{2} + 10 x + 24) = \frac{1}{2}$ 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 $b^{y} = x$ .

$9^{\frac{1}{2}} = x^{2} + 10 x + 24$

Solve for $x$ .

Rewrite the equation as $x^{2} + 10 x + 24 = 9^{\frac{1}{2}}$ .

$x^{2} + 10 x + 24 = 9^{\frac{1}{2}}$

Simplify $9^{\frac{1}{2}}$ .

Rewrite.

$x^{2} + 10 x + 24 = 0 + 0 + 9^{\frac{1}{2}}$

Simplify the expression.

Rewrite $9$ as $3^{2}$ .

$x^{2} + 10 x + 24 = {(3^{2})}^{\frac{1}{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2} + 10 x + 24 = 3^{2 (\frac{1}{2})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} + 10 x + 24 = 3^{2 (\frac{1}{2})}$

Rewrite the expression.

$x^{2} + 10 x + 24 = 3^{1}$

Evaluate the exponent.

$x^{2} + 10 x + 24 = 3$

Subtract $3$ from both sides of the equation.

$x^{2} + 10 x + 24 - 3 = 0$

Subtract $3$ from $24$ .

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

Factor $x^{2} + 10 x + 21$ 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 $21$ and whose sum is $10$ .

$3, 7$

Write the factored form using these integers.

$(x + 3) (x + 7) = 0$

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

$x + 3 = 0$

$x + 7 = 0$

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 $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x + 3) (x + 7) = 0$ true.

$x = - 3, - 7$

Exclude the solutions that do not make $\log_{9} (x + 6) + \log_{9} (x + 4) = \frac{1}{2}$ true.

$x = - 3$

Solve for x log base 9 of x+6+ log base 9 of x+4=1/2

Related Questions

Support LCMGCF.com by disabling your adblocker.