Finite Math Examples

5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5

$5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5

$5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5$

Step 2

Simplify the left side.

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Step 2.1

Simplify

5 \log_{2} (x) - \log_{2} (2 x^{3})

$5 \log_{2} (x) - \log_{2} (2 x^{3})$ .

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Step 2.1.1

Simplify

5 \log_{2} (x)

$5 \log_{2} (x)$ by moving

5

$5$ inside the logarithm.

\log_{2} (x^{5}) - \log_{2} (2 x^{3}) = 5

$\log_{2} (x^{5}) - \log_{2} (2 x^{3}) = 5$

Step 2.1.2

Use the quotient property of logarithms,

\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

\log_{2} (\frac{x^{5}}{2 x^{3}}) = 5

$\log_{2} (\frac{x^{5}}{2 x^{3}}) = 5$

Step 2.1.3

Reduce the expression

\frac{x^{5}}{2 x^{3}}

$\frac{x^{5}}{2 x^{3}}$ by cancelling the common factors.

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Step 2.1.3.1

Factor

x^{3}

$x^{3}$ out of

x^{5}

$x^{5}$ .

\log_{2} (\frac{x^{3} x^{2}}{2 x^{3}}) = 5

$\log_{2} (\frac{x^{3} x^{2}}{2 x^{3}}) = 5$

Step 2.1.3.2

Factor

x^{3}

$x^{3}$ out of

2 x^{3}

$2 x^{3}$ .

\log_{2} (\frac{x^{3} x^{2}}{x^{3} \cdot 2}) = 5

$\log_{2} (\frac{x^{3} x^{2}}{x^{3} \cdot 2}) = 5$

Step 2.1.3.3

Cancel the common factor.

\log_{2} (\frac{x^{3} x^{2}}{x^{3} \cdot 2}) = 5

$\log_{2} (\frac{x^{3} x^{2}}{x^{3} \cdot 2}) = 5$

Step 2.1.3.4

Rewrite the expression.

\log_{2} (\frac{x^{2}}{2}) = 5

$\log_{2} (\frac{x^{2}}{2}) = 5$

\log_{2} (\frac{x^{2}}{2}) = 5

$\log_{2} (\frac{x^{2}}{2}) = 5$

Step 2.1.4

Rewrite

\frac{x^{2}}{2}

$\frac{x^{2}}{2}$ as

x^{2} \cdot 2^{- 1}

$x^{2} \cdot 2^{- 1}$ .

\log_{2} (x^{2} \cdot 2^{- 1}) = 5

$\log_{2} (x^{2} \cdot 2^{- 1}) = 5$

Step 2.1.5

Rewrite

\log_{2} (x^{2} \cdot 2^{- 1})

$\log_{2} (x^{2} \cdot 2^{- 1})$ as

\log_{2} (x^{2}) + \log_{2} (2^{- 1})

$\log_{2} (x^{2}) + \log_{2} (2^{- 1})$ .

\log_{2} (x^{2}) + \log_{2} (2^{- 1}) = 5

$\log_{2} (x^{2}) + \log_{2} (2^{- 1}) = 5$

Step 2.1.6

Use logarithm rules to move

- 1

$- 1$ out of the exponent.

\log_{2} (x^{2}) - \log_{2} (2) = 5

$\log_{2} (x^{2}) - \log_{2} (2) = 5$

Step 2.1.7

Logarithm base

2

$2$ of

2

$2$ is

1

$1$ .

\log_{2} (x^{2}) - 1 \cdot 1 = 5

$\log_{2} (x^{2}) - 1 \cdot 1 = 5$

Step 2.1.8

Multiply

- 1

$- 1$ by

1

$1$ .

\log_{2} (x^{2}) - 1 = 5

$\log_{2} (x^{2}) - 1 = 5$

\log_{2} (x^{2}) - 1 = 5

$\log_{2} (x^{2}) - 1 = 5$

\log_{2} (x^{2}) - 1 = 5

$\log_{2} (x^{2}) - 1 = 5$

Step 3

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 3.1

Add

1

$1$ to both sides of the equation.

\log_{2} (x^{2}) = 5 + 1

$\log_{2} (x^{2}) = 5 + 1$

Step 3.2

Add

5

$5$ and

1

$1$ .

\log_{2} (x^{2}) = 6

$\log_{2} (x^{2}) = 6$

\log_{2} (x^{2}) = 6

$\log_{2} (x^{2}) = 6$

Step 4

Write in exponential form.

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Step 4.1

For logarithmic equations,

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ such that

x > 0

$x > 0$ ,

b > 0

$b > 0$ , and

b \neq 1

$b \neq 1$ . In this case,

b = 2

$b = 2$ ,

x = x^{2}

$x = x^{2}$ , and

y = 6

$y = 6$ .

b = 2

$b = 2$

x = x^{2}

$x = x^{2}$

y = 6

$y = 6$

Step 4.2

Substitute the values of

b

$b$ ,

x

$x$ , and

y

$y$ into the equation

b^{y} = x

$b^{y} = x$ .

2^{6} = x^{2}

$2^{6} = x^{2}$

2^{6} = x^{2}

$2^{6} = x^{2}$

Step 5

Solve for

x

$x$ .

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Step 5.1

Rewrite the equation as

x^{2} = 2^{6}

$x^{2} = 2^{6}$ .

x^{2} = 2^{6}

$x^{2} = 2^{6}$

Step 5.2

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

x = \pm \sqrt{2^{6}}

$x = \pm \sqrt{2^{6}}$

Step 5.3

Simplify

\pm \sqrt{2^{6}}

$\pm \sqrt{2^{6}}$ .

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Step 5.3.1

Raise

2

$2$ to the power of

6

$6$ .

x = \pm \sqrt{64}

$x = \pm \sqrt{64}$

Step 5.3.2

Rewrite

64

$64$ as

8^{2}

$8^{2}$ .

x = \pm \sqrt{8^{2}}

$x = \pm \sqrt{8^{2}}$

Step 5.3.3

Pull terms out from under the radical, assuming positive real numbers.

x = \pm 8

$x = \pm 8$

x = \pm 8

$x = \pm 8$

Step 5.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.4.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = 8

$x = 8$

Step 5.4.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - 8

$x = - 8$

Step 5.4.3

The complete solution is the result of both the positive and negative portions of the solution.

x = 8, - 8

$x = 8, - 8$

x = 8, - 8

$x = 8, - 8$

x = 8, - 8

$x = 8, - 8$

Step 6

Exclude the solutions that do not make

5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5

$5 \log_{2} (x) - \log_{2} (2 x^{3}) = 5$ true.

x = 8

$x = 8$