Solve for x log base 4 of x+6+ log base 4 of x=2

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

Simplify the left side.

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

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

Apply the distributive property.

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

Multiply $x$ by $x$ .

$\log_{4} (x^{2} + 6 x) = 2$

Rewrite $\log_{4} (x^{2} + 6 x) = 2$ 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 $b^{y} = x$ .

$4^{2} = x^{2} + 6 x$

Solve for $x$ .

Rewrite the equation as $x^{2} + 6 x = 4^{2}$ .

$x^{2} + 6 x = 4^{2}$

Raise $4$ to the power of $2$ .

$x^{2} + 6 x = 16$

Subtract $16$ from both sides of the equation.

$x^{2} + 6 x - 16 = 0$

Factor $x^{2} + 6 x - 16$ 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 $6$ .

$- 2, 8$

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 $0$ .

$x - 2 = 0$

Add $2$ to 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 $\log_{4} (x + 6) + \log_{4} (x) = 2$ true.

$x = 2$

Solve for x log base 4 of x+6+ log base 4 of x=2

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