Solve for x log base 9 of -11x+2 = log base 9 of x^2+30

$\log_{9} (- 11 x + 2) = \log_{9} (x^{2} + 30)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$- 11 x + 2 = x^{2} + 30$

Step 2

Solve for $x$ .

Subtract $x^{2}$ from both sides of the equation.

$- 11 x + 2 - x^{2} = 30$

Subtract $30$ from both sides of the equation.

$- 11 x + 2 - x^{2} - 30 = 0$

Subtract $30$ from $2$ .

$- 11 x - x^{2} - 28 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 11 x - x^{2} - 28$ .

Reorder $- 11 x$ and $- x^{2}$ .

$- x^{2} - 11 x - 28 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 11 x - 28 = 0$

Factor $- 1$ out of $- 11 x$ .

$- (x^{2}) - (11 x) - 28 = 0$

Rewrite $- 28$ as $- 1 (28)$ .

$- (x^{2}) - (11 x) - 1 \cdot 28 = 0$

Factor $- 1$ out of $- (x^{2}) - (11 x)$ .

$- (x^{2} + 11 x) - 1 \cdot 28 = 0$

Factor $- 1$ out of $- (x^{2} + 11 x) - 1 (28)$ .

$- (x^{2} + 11 x + 28) = 0$

Factor.

Factor $x^{2} + 11 x + 28$ 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 $28$ and whose sum is $11$ .

$4, 7$

Write the factored form using these integers.

$- ((x + 4) (x + 7)) = 0$

Remove unnecessary parentheses.

$- (x + 4) (x + 7) = 0$

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

$x + 4 = 0$

$x + 7 = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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 $- (x + 4) (x + 7) = 0$ true.

$x = - 4, - 7$

Solve for x log base 9 of -11x+2 = log base 9 of x^2+30

Related Questions

Support LCMGCF.com by disabling your adblocker.