Trigonometry Examples

$\log_{12} (x) + \log_{12} (x - 1) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{12} (x (x - 1)) = 1$

Step 1.2

Simplify by multiplying through.

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Step 1.2.1

Apply the distributive property.

$\log_{12} (x \cdot x + x \cdot - 1) = 1$

Step 1.2.2

Simplify the expression.

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Step 1.2.2.1

Multiply $x$ by $x$ .

$\log_{12} (x^{2} + x \cdot - 1) = 1$

Step 1.2.2.2

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

$\log_{12} (x^{2} - 1 \cdot x) = 1$

Step 1.3

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

$\log_{12} (x^{2} - x) = 1$

Step 2

Rewrite $\log_{12} (x^{2} - x) = 1$ 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 $b^{y} = x$ .

$12^{1} = x^{2} - x$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - x = 12^{1}$ .

$x^{2} - x = 12$

Step 3.2

Subtract $12$ from both sides of the equation.

$x^{2} - x - 12 = 0$

Step 3.3

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 3.3.2

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

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 $0$ .

$x - 4 = 0$

$x + 3 = 0$

Step 3.5

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

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Step 3.5.1

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

$x - 4 = 0$

Step 3.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.6

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 3.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.7

The final solution is all the values that make $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Step 4

Exclude the solutions that do not make $\log_{12} (x) + \log_{12} (x - 1) = 1$ true.

$x = 4$