Algebra Examples

$\log_{4} (x^{2} - x) = 1 + \log_{4} (5)$

Step 1

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

$\log_{4} (x^{2} - x) - \log_{4} (5) = 1$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{4} (\frac{x^{2} - x}{5}) = 1$

Step 3

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

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Step 3.1

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

$\log_{4} (\frac{x \cdot x - x}{5}) = 1$

Step 3.2

Factor $x$ out of $- x$ .

$\log_{4} (\frac{x \cdot x + x \cdot - 1}{5}) = 1$

Step 3.3

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

$\log_{4} (\frac{x (x - 1)}{5}) = 1$

Step 4

Rewrite $\log_{4} (\frac{x (x - 1)}{5}) = 1$ 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 $b^{y} = x$ .

$4^{1} = \frac{x (x - 1)}{5}$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $\frac{x (x - 1)}{5} = 4^{1}$ .

$\frac{x (x - 1)}{5} = 4$

Step 5.2

Multiply both sides of the equation by $5$ .

$5 \frac{x (x - 1)}{5} = 5 \cdot 4^{1}$

Step 5.3

Simplify both sides of the equation.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Simplify $5 \frac{x (x - 1)}{5}$ .

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Step 5.3.1.1.1

Simplify terms.

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Step 5.3.1.1.1.1

Cancel the common factor of $5$ .

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Step 5.3.1.1.1.1.1

Cancel the common factor.

$5 \frac{x (x - 1)}{5} = 5 \cdot 4^{1}$

Step 5.3.1.1.1.1.2

Rewrite the expression.

$x (x - 1) = 5 \cdot 4^{1}$

Step 5.3.1.1.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 1 = 5 \cdot 4^{1}$

Step 5.3.1.1.1.3

Simplify the expression.

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Step 5.3.1.1.1.3.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 = 5 \cdot 4^{1}$

Step 5.3.1.1.1.3.2

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x = 5 \cdot 4^{1}$

Step 5.3.1.1.2

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x = 5 \cdot 4^{1}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $5 \cdot 4^{1}$ .

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Step 5.3.2.1.1

Evaluate the exponent.

$x^{2} - x = 5 \cdot 4$

Step 5.3.2.1.2

Multiply $5$ by $4$ .

$x^{2} - x = 20$

Step 5.4

Subtract $20$ from both sides of the equation.

$x^{2} - x - 20 = 0$

Step 5.5

Factor $x^{2} - x - 20$ using the AC method.

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Step 5.5.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 5.5.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 5.6

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

$x - 5 = 0$

$x + 4 = 0$

Step 5.7

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

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Step 5.7.1

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

$x - 5 = 0$

Step 5.7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 5.8

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 5.8.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 5.8.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5.9

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$