Solve the Inequality for x log base 3 of 1-x> = log base 3 of x+16-x^2

$\log_{3} (1 - x) \geq \log_{3} (x + 16 - x^{2})$

Convert the inequality to an equality.

$\log_{3} (1 - x) = \log_{3} (x + 16 - x^{2})$

Solve the equation.

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

$1 - x = x + 16 - x^{2}$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x + 16 - x^{2} = 1 - x$

Move all terms containing $x$ to the left side of the equation.

Add $x$ to both sides of the equation.

$x + 16 - x^{2} + x = 1$

Add $x$ and $x$ .

$2 x + 16 - x^{2} = 1$

Subtract $1$ from both sides of the equation.

$2 x + 16 - x^{2} - 1 = 0$

Subtract $1$ from $16$ .

$2 x - x^{2} + 15 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $2 x - x^{2} + 15$ .

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

$- x^{2} + 2 x + 15 = 0$

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

$- (x^{2}) + 2 x + 15 = 0$

Factor $- 1$ out of $2 x$ .

$- (x^{2}) - (- 2 x) + 15 = 0$

Rewrite $15$ as $- 1 (- 15)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 15 = 0$

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

$- (x^{2} - 2 x) - 1 \cdot - 15 = 0$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 15)$ .

$- (x^{2} - 2 x - 15) = 0$

Factor.

Factor $x^{2} - 2 x - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$- ((x - 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$- (x - 5) (x + 3) = 0$

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

$x - 5 = 0$

$x + 3 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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 $- (x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Find the domain of $\log_{3} (\frac{1 - x}{- x^{2} + x + 16})$ .

Set the argument in $\log_{3} (\frac{1 - x}{- x^{2} + x + 16})$ greater than $0$ to find where the expression is defined.

$\frac{1 - x}{- x^{2} + x + 16} > 0$

Solve for $x$ .

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$1 - x = 0$

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

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 1$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 16)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 16}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot 16}}{2 \cdot - 1}$

Multiply $4$ by $16$ .

$x = \frac{- 1 \pm \sqrt{1 + 64}}{2 \cdot - 1}$

Add $1$ and $64$ .

$x = \frac{- 1 \pm \sqrt{65}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{65}}{- 2}$

Simplify $\frac{- 1 \pm \sqrt{65}}{- 2}$ .

$x = \frac{1 \pm \sqrt{65}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{65}}{2}, \frac{1 - \sqrt{65}}{2}$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1$

$x = \frac{1 + \sqrt{65}}{2}, \frac{1 - \sqrt{65}}{2}$

Consolidate the solutions.

$x = 1, \frac{1 + \sqrt{65}}{2}, \frac{1 - \sqrt{65}}{2}$

Find the domain of $\frac{1 - x}{- x^{2} + x + 16}$ .

Set the denominator in $\frac{1 - x}{- x^{2} + x + 16}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 1$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 16)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 16}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot 16}}{2 \cdot - 1}$

Multiply $4$ by $16$ .

$x = \frac{- 1 \pm \sqrt{1 + 64}}{2 \cdot - 1}$

Add $1$ and $64$ .

$x = \frac{- 1 \pm \sqrt{65}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{65}}{- 2}$

Simplify $\frac{- 1 \pm \sqrt{65}}{- 2}$ .

$x = \frac{1 \pm \sqrt{65}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{65}}{2}, \frac{1 - \sqrt{65}}{2}$

The domain is all values of $x$ that make the expression defined.

$(- \infty, \frac{1 - \sqrt{65}}{2}) \cup (\frac{1 - \sqrt{65}}{2}, \frac{1 + \sqrt{65}}{2}) \cup (\frac{1 + \sqrt{65}}{2}, \infty)$

Use each root to create test intervals.

$x < \frac{1 - \sqrt{65}}{2}$

$\frac{1 - \sqrt{65}}{2} < x < 1$

$1 < x < \frac{1 + \sqrt{65}}{2}$

$x > \frac{1 + \sqrt{65}}{2}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < \frac{1 - \sqrt{65}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1 - \sqrt{65}}{2}$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\frac{1 - (- 6)}{- {(- 6)}^{2} - 6 + 16} > 0$

The left side $- 0.2\overline{692307}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $\frac{1 - \sqrt{65}}{2} < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1 - \sqrt{65}}{2} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{1 - (0)}{- {(0)}^{2} + 0 + 16} > 0$

The left side $0.0625$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < \frac{1 + \sqrt{65}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < \frac{1 + \sqrt{65}}{2}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$\frac{1 - (3)}{- {(3)}^{2} + 3 + 16} > 0$

The left side $- 0.2$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1 + \sqrt{65}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1 + \sqrt{65}}{2}$ and see if this value makes the original inequality true.

$x = 7$

Replace $x$ with $7$ in the original inequality.

$\frac{1 - (7)}{- {(7)}^{2} + 7 + 16} > 0$

The left side $0.\overline{230769}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{1 - \sqrt{65}}{2}$ False

$\frac{1 - \sqrt{65}}{2} < x < 1$ True

$1 < x < \frac{1 + \sqrt{65}}{2}$ False

$x > \frac{1 + \sqrt{65}}{2}$ True

$x < \frac{1 - \sqrt{65}}{2}$ False

$\frac{1 - \sqrt{65}}{2} < x < 1$ True

$1 < x < \frac{1 + \sqrt{65}}{2}$ False

$x > \frac{1 + \sqrt{65}}{2}$ True

The solution consists of all of the true intervals.

$\frac{1 - \sqrt{65}}{2} < x < 1$ or $x > \frac{1 + \sqrt{65}}{2}$

Set the denominator in $\frac{1 - x}{- x^{2} + x + 16}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 1$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 16)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 16}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot 16}}{2 \cdot - 1}$

Multiply $4$ by $16$ .

$x = \frac{- 1 \pm \sqrt{1 + 64}}{2 \cdot - 1}$

Add $1$ and $64$ .

$x = \frac{- 1 \pm \sqrt{65}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{65}}{- 2}$

Simplify $\frac{- 1 \pm \sqrt{65}}{- 2}$ .

$x = \frac{1 \pm \sqrt{65}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{65}}{2}, \frac{1 - \sqrt{65}}{2}$

The domain is all values of $x$ that make the expression defined.

$(\frac{1 - \sqrt{65}}{2}, 1) \cup (\frac{1 + \sqrt{65}}{2}, \infty)$

Use each root to create test intervals.

$x < \frac{1 - \sqrt{65}}{2}$

$\frac{1 - \sqrt{65}}{2} < x < - 3$

$- 3 < x < 1$

$1 < x < \frac{1 + \sqrt{65}}{2}$

$\frac{1 + \sqrt{65}}{2} < x < 5$

$x > 5$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < \frac{1 - \sqrt{65}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1 - \sqrt{65}}{2}$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\log_{3} (1 - (- 6)) \geq \log_{3} ((- 6) + 16 - {(- 6)}^{2})$

Determine if the inequality is true.

The equation cannot be solved because it is undefined.

$Undefined$

The right side has no solution, which means that the given statement is false.

False

Test a value on the interval $\frac{1 - \sqrt{65}}{2} < x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1 - \sqrt{65}}{2} < x < - 3$ and see if this value makes the original inequality true.

$x = - 3.27$

Replace $x$ with $- 3.27$ in the original inequality.

$\log_{3} (1 - (- 3.27)) \geq \log_{3} ((- 3.27) + 16 - {(- 3.27)}^{2})$

The left side $1.32131584$ is greater than the right side $0.64765999$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\log_{3} (1 - (0)) \geq \log_{3} ((0) + 16 - {(0)}^{2})$

The left side $0$ is less than the right side $2.52371901$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < \frac{1 + \sqrt{65}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < \frac{1 + \sqrt{65}}{2}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$\log_{3} (1 - (3)) \geq \log_{3} ((3) + 16 - {(3)}^{2})$

Determine if the inequality is true.

The equation cannot be solved because it is undefined.

$Undefined$

The left side has no solution, which means that the given statement is false.

False

Test a value on the interval $\frac{1 + \sqrt{65}}{2} < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1 + \sqrt{65}}{2} < x < 5$ and see if this value makes the original inequality true.

$x = 4.77$

Replace $x$ with $4.77$ in the original inequality.

$\log_{3} (1 - (4.77)) \geq \log_{3} ((4.77) + 16 - {(4.77)}^{2})$

Determine if the inequality is true.

The equation cannot be solved because it is undefined.

$Undefined$

The left side has no solution, which means that the given statement is false.

False

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\log_{3} (1 - (8)) \geq \log_{3} ((8) + 16 - {(8)}^{2})$

Determine if the inequality is true.

The equation cannot be solved because it is undefined.

$Undefined$

The left side has no solution, which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{1 - \sqrt{65}}{2}$ False

$\frac{1 - \sqrt{65}}{2} < x < - 3$ True

$- 3 < x < 1$ False

$1 < x < \frac{1 + \sqrt{65}}{2}$ False

$\frac{1 + \sqrt{65}}{2} < x < 5$ False

$x > 5$ False

$x < \frac{1 - \sqrt{65}}{2}$ False

$\frac{1 - \sqrt{65}}{2} < x < - 3$ True

$- 3 < x < 1$ False

$1 < x < \frac{1 + \sqrt{65}}{2}$ False

$\frac{1 + \sqrt{65}}{2} < x < 5$ False

$x > 5$ False

The solution consists of all of the true intervals.

$\frac{1 - \sqrt{65}}{2} < x \leq - 3$

The result can be shown in multiple forms.

Inequality Form:

$\frac{1 - \sqrt{65}}{2} < x \leq - 3$

Interval Notation:

$(\frac{1 - \sqrt{65}}{2}, - 3]$

Solve the Inequality for x log base 3 of 1-x> = log base 3 of x+16-x^2

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