Precalculus Examples

$\log (x + 9) = 1 - \log (x)$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $1$ from both sides of the equation.

$\log (x + 9) - 1 = - \log (x)$

Step 1.2

Add $\log (x)$ to both sides of the equation.

$\log (x + 9) - 1 + \log (x) = 0$

Step 2

Simplify $\log (x + 9) - 1 + \log (x)$ .

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Step 2.1

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

$\log ((x + 9) x) - 1 = 0$

Step 2.2

Simplify each term.

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Step 2.2.1

Apply the distributive property.

$\log (x \cdot x + 9 x) - 1 = 0$

Step 2.2.2

Multiply $x$ by $x$ .

$\log (x^{2} + 9 x) - 1 = 0$

Step 3

Add $1$ to both sides of the equation.

$\log (x^{2} + 9 x) = 1$

Step 4

Rewrite $\log (x^{2} + 9 x) = 1$ 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 $b^{y} = x$ .

$10^{1} = x^{2} + 9 x$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $x^{2} + 9 x = 10^{1}$ .

$x^{2} + 9 x = 10$

Step 5.2

Subtract $10$ from both sides of the equation.

$x^{2} + 9 x - 10 = 0$

Step 5.3

Factor $x^{2} + 9 x - 10$ using the AC method.

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Step 5.3.1

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 $- 10$ and whose sum is $9$ .

$- 1, 10$

Step 5.3.2

Write the factored form using these integers.

$(x - 1) (x + 10) = 0$

Step 5.4

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

$x - 1 = 0$

$x + 10 = 0$

Step 5.5

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

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Step 5.5.1

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

$x - 1 = 0$

Step 5.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.6

Set $x + 10$ equal to $0$ and solve for $x$ .

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Step 5.6.1

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Step 5.6.2

Subtract $10$ from both sides of the equation.

$x = - 10$

Step 5.7

The final solution is all the values that make $(x - 1) (x + 10) = 0$ true.

$x = 1, - 10$