Solve the Inequality for x log of x+ log of x-3< = log of 10

$\log (x) + \log (x - 3) \leq \log (10)$

Convert the inequality to an equality.

$\log (x) + \log (x - 3) = \log (10)$

Solve the equation.

Simplify the left side.

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

$\log (x (x - 3)) = \log (10)$

Apply the distributive property.

$\log (x \cdot x + x \cdot - 3) = \log (10)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 3) = \log (10)$

Move $- 3$ to the left of $x$ .

$\log (x^{2} - 3 x) = \log (10)$

Logarithm base $10$ of $10$ is $1$ .

$\log (x^{2} - 3 x) = 1$

Rewrite $\log (x^{2} - 3 x) = 1$ 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 $b^{y} = x$ .

$10^{1} = x^{2} - 3 x$

Solve for $x$ .

Rewrite the equation as $x^{2} - 3 x = 10^{1}$ .

$x^{2} - 3 x = 10$

Subtract $10$ from both sides of the equation.

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

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x - 5) (x + 2) = 0$

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

$x - 5 = 0$

$x + 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

Find the domain of $\log (x^{2} - 3 x) - 1$ .

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

$x^{2} - 3 x > 0$

Solve for $x$ .

Convert the inequality to an equation.

$x^{2} - 3 x = 0$

Factor $x$ out of $x^{2} - 3 x$ .

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

$x \cdot x - 3 x = 0$

Factor $x$ out of $- 3 x$ .

$x \cdot x + x \cdot - 3 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 3$ .

$x (x - 3) = 0$

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

$x = 0$

$x - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 0, 3$

Use each root to create test intervals.

$x < 0$

$0 < x < 3$

$x > 3$

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 $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

${(- 2)}^{2} - 3 \cdot - 2 > 0$

The left side $10$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

${(2)}^{2} - 3 \cdot 2 > 0$

The left side $- 2$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 6$

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

${(6)}^{2} - 3 \cdot 6 > 0$

The left side $18$ 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.

$x < 0$ True

$0 < x < 3$ False

$x > 3$ True

$x < 0$ True

$0 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x < 0$ or $x > 3$

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

$(- \infty, 0) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 3$

$3 < x < 5$

$x > 5$

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 $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\log (- 4) + \log ((- 4) - 3) \leq \log (10)$

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 $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$\log (- 1) + \log ((- 1) - 3) \leq \log (10)$

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 $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\log (2) + \log ((2) - 3) \leq \log (10)$

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 $3 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\log (4) + \log ((4) - 3) \leq \log (10)$

The left side $0.60205999$ is less than the right side $1$ , which means that the given statement is always true.

True

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.

$\log (8) + \log ((8) - 3) \leq \log (10)$

The left side $1.60205999$ is greater than the right side $1$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < 0$ False

$0 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

$x < - 2$ False

$- 2 < x < 0$ False

$0 < x < 3$ False

$3 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$3 < x \leq 5$

The result can be shown in multiple forms.

Inequality Form:

$3 < x \leq 5$

Interval Notation:

$(3, 5]$

Solve the Inequality for x log of x+ log of x-3< = log of 10

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