Precalculus Examples

$\log (x) + \log (x - 30) = 3$

Step 1

Simplify the left side.

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Step 1.1

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

$\log (x (x - 30)) = 3$

Step 1.2

Apply the distributive property.

$\log (x \cdot x + x \cdot - 30) = 3$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 30) = 3$

Step 1.3.2

Move $- 30$ to the left of $x$ .

$\log (x^{2} - 30 x) = 3$

Step 2

Rewrite $\log (x^{2} - 30 x) = 3$ 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 $b^{y} = x$ .

$10^{3} = x^{2} - 30 x$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - 30 x = 10^{3}$ .

$x^{2} - 30 x = 10^{3}$

Step 3.2

Raise $10$ to the power of $3$ .

$x^{2} - 30 x = 1000$

Step 3.3

Subtract $1000$ from both sides of the equation.

$x^{2} - 30 x - 1000 = 0$

Step 3.4

Factor $x^{2} - 30 x - 1000$ using the AC method.

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Step 3.4.1

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 $- 1000$ and whose sum is $- 30$ .

$- 50, 20$

Step 3.4.2

Write the factored form using these integers.

$(x - 50) (x + 20) = 0$

Step 3.5

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

$x - 50 = 0$

$x + 20 = 0$

Step 3.6

Set $x - 50$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 50$ equal to $0$ .

$x - 50 = 0$

Step 3.6.2

Add $50$ to both sides of the equation.

$x = 50$

Step 3.7

Set $x + 20$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x + 20$ equal to $0$ .

$x + 20 = 0$

Step 3.7.2

Subtract $20$ from both sides of the equation.

$x = - 20$

Step 3.8

The final solution is all the values that make $(x - 50) (x + 20) = 0$ true.

$x = 50, - 20$

Step 4

Exclude the solutions that do not make $\log (x) + \log (x - 30) = 3$ true.

$x = 50$