Precalculus Examples

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

Step 1

Simplify the left side.

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Step 1.1

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

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

Step 1.2

Apply the distributive property.

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

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

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

Step 1.3.2

Move $- 4$ to the left of $x$ .

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

Step 2

Rewrite $\log_{5} (x^{2} - 4 x) = 1$ 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 $b^{y} = x$ .

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

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - 4 x = 5^{1}$ .

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

Step 3.2

Subtract $5$ from both sides of the equation.

$x^{2} - 4 x - 5 = 0$

Step 3.3

Factor $x^{2} - 4 x - 5$ using the AC method.

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Step 3.3.1

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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 3.3.2

Write the factored form using these integers.

$(x - 5) (x + 1) = 0$

Step 3.4

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

$x - 5 = 0$

$x + 1 = 0$

Step 3.5

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

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Step 3.5.1

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

$x - 5 = 0$

Step 3.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 3.6

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.7

The final solution is all the values that make $(x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Step 4

Exclude the solutions that do not make $\log_{5} (x) + \log_{5} (x - 4) = 1$ true.

$x = 5$