Algebra Examples

$- 3 \log_{5} (x) + 6 \leq 9$

Step 1

Convert the inequality to an equality.

$- 3 \log_{5} (x) + 6 = 9$

Step 2

Solve the equation.

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Step 2.1

Move all terms not containing $\log_{5} (x)$ to the right side of the equation.

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Step 2.1.1

Subtract $6$ from both sides of the equation.

$- 3 \log_{5} (x) = 9 - 6$

Step 2.1.2

Subtract $6$ from $9$ .

$- 3 \log_{5} (x) = 3$

Step 2.2

Divide each term in $- 3 \log_{5} (x) = 3$ by $- 3$ and simplify.

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Step 2.2.1

Divide each term in $- 3 \log_{5} (x) = 3$ by $- 3$ .

$\frac{- 3 \log_{5} (x)}{- 3} = \frac{3}{- 3}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 3$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 3 \log_{5} (x)}{- 3} = \frac{3}{- 3}$

Step 2.2.2.1.2

Divide $\log_{5} (x)$ by $1$ .

$\log_{5} (x) = \frac{3}{- 3}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $3$ by $- 3$ .

$\log_{5} (x) = - 1$

Step 2.3

Rewrite $\log_{5} (x) = - 1$ 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 $b^{y} = x$ .

$5^{- 1} = x$

Step 2.4

Solve for $x$ .

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Step 2.4.1

Rewrite the equation as $x = 5^{- 1}$ .

$x = 5^{- 1}$

Step 2.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{5}$

Step 3

Find the domain of $- \log_{5} (x^{3}) - 3$ .

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Step 3.1

Set the argument in $\log_{5} (x^{3})$ greater than $0$ to find where the expression is defined.

$x^{3} > 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt[3]{x^{3}} > \sqrt[3]{0}$

Step 3.2.2

Simplify the equation.

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Pull terms out from under the radical.

$x > \sqrt[3]{0}$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify $\sqrt[3]{0}$ .

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Step 3.2.2.2.1.1

Rewrite $0$ as $0^{3}$ .

$x > \sqrt[3]{0^{3}}$

Step 3.2.2.2.1.2

Pull terms out from under the radical.

$x > 0$

Step 3.3

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

$(0, \infty)$

Step 4

The solution consists of all of the true intervals.

$x \geq \frac{1}{5}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \geq \frac{1}{5}$

Interval Notation:

$[\frac{1}{5}, \infty)$

Step 6