Solve for x log base 3 of x+ log base 3 of x-6=3

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

Step 1

Simplify the left side.

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

$\log_{3} (x (x - 6)) = 3$

Apply the distributive property.

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

Simplify the expression.

Multiply $x$ by $x$ .

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

Move $- 6$ to the left of $x$ .

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

Step 2

Rewrite $\log_{3} (x^{2} - 6 x) = 3$ 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 $b^{y} = x$ .

$3^{3} = x^{2} - 6 x$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 6 x = 3^{3}$ .

$x^{2} - 6 x = 3^{3}$

Raise $3$ to the power of $3$ .

$x^{2} - 6 x = 27$

Subtract $27$ from both sides of the equation.

$x^{2} - 6 x - 27 = 0$

Factor $x^{2} - 6 x - 27$ 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 $- 6$ .

$- 9, 3$

Write the factored form using these integers.

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

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

$x - 9 = 0$

$x + 3 = 0$

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

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

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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.

$x = - 3$

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

$x = 9, - 3$

Step 4

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

$x = 9$

Solve for x log base 3 of x+ log base 3 of x-6=3

Related Questions

Support LCMGCF.com by disabling your adblocker.