Precalculus Examples

$\log (x - 15) = 2 - \log (x)$

Step 1

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

$\log (x - 15) + \log (x) = 2$

Step 2

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

$\log ((x - 15) x) = 2$

Step 3

Apply the distributive property.

$\log (x \cdot x - 15 x) = 2$

Step 4

Multiply $x$ by $x$ .

$\log (x^{2} - 15 x) = 2$

Step 5

Rewrite $\log (x^{2} - 15 x) = 2$ 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 $b^{y} = x$ .

$10^{2} = x^{2} - 15 x$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $x^{2} - 15 x = 10^{2}$ .

$x^{2} - 15 x = 10^{2}$

Step 6.2

Raise $10$ to the power of $2$ .

$x^{2} - 15 x = 100$

Step 6.3

Subtract $100$ from both sides of the equation.

$x^{2} - 15 x - 100 = 0$

Step 6.4

Factor $x^{2} - 15 x - 100$ using the AC method.

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Step 6.4.1

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 $- 100$ and whose sum is $- 15$ .

$- 20, 5$

Step 6.4.2

Write the factored form using these integers.

$(x - 20) (x + 5) = 0$

Step 6.5

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

$x - 20 = 0$

$x + 5 = 0$

Step 6.6

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

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Step 6.6.1

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

$x - 20 = 0$

Step 6.6.2

Add $20$ to both sides of the equation.

$x = 20$

Step 6.7

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

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Step 6.7.1

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

$x + 5 = 0$

Step 6.7.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6.8

The final solution is all the values that make $(x - 20) (x + 5) = 0$ true.

$x = 20, - 5$

Step 7

Exclude the solutions that do not make $\log (x - 15) = 2 - \log (x)$ true.

$x = 20$