Solve for a log base 11 of a+7<1

$\log_{11} (a + 7) < 1$

Convert the inequality to an equality.

$\log_{11} (a + 7) = 1$

Solve the equation.

Rewrite $\log_{11} (a + 7) = 1$ 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 $b^{y} = x$ .

$11^{1} = a + 7$

Solve for $a$ .

Rewrite the equation as $a + 7 = 11^{1}$ .

$a + 7 = 11$

Move all terms not containing $a$ to the right side of the equation.

Subtract $7$ from both sides of the equation.

$a = 11 - 7$

Subtract $7$ from $11$ .

$a = 4$

Find the domain of $\log_{11} (a + 7) - 1$ .

Set the argument in $\log_{11} (a + 7)$ greater than $0$ to find where the expression is defined.

$a + 7 > 0$

Subtract $7$ from both sides of the inequality.

$a > - 7$

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

$(- 7, \infty)$

Use each root to create test intervals.

$a < - 7$

$- 7 < a < 4$

$a > 4$

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 $a < - 7$ to see if it makes the inequality true.

Choose a value on the interval $a < - 7$ and see if this value makes the original inequality true.

$a = - 10$

Replace $a$ with $- 10$ in the original inequality.

$\log_{11} ((- 10) + 7) < 1$

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 $- 7 < a < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < a < 4$ and see if this value makes the original inequality true.

$a = 0$

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

$\log_{11} ((0) + 7) < 1$

The left side $0.81150756$ is less than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $a > 4$ to see if it makes the inequality true.

Choose a value on the interval $a > 4$ and see if this value makes the original inequality true.

$a = 6$

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

$\log_{11} ((6) + 7) < 1$

The left side $1.06966696$ is not less 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.

$a < - 7$ False

$- 7 < a < 4$ True

$a > 4$ False

$a < - 7$ False

$- 7 < a < 4$ True

$a > 4$ False

The solution consists of all of the true intervals.

$- 7 < a < 4$

The result can be shown in multiple forms.

Inequality Form:

$- 7 < a < 4$

Interval Notation:

$(- 7, 4)$

Solve for a log base 11 of a+7<1

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