Finite Math Examples

$\ln (x) + \ln (x + 1) = \ln (6)$

Step 1

Subtract $\ln (6)$ from both sides of the equation.

$\ln (x) + \ln (x + 1) - \ln (6) = 0$

Step 2

Simplify $\ln (x) + \ln (x + 1) - \ln (6)$ .

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Step 2.1

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

$\ln (x (x + 1)) - \ln (6) = 0$

Step 2.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{x (x + 1)}{6}) = 0$

Step 3

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{x (x + 1)}{6})} = e^{0}$

Step 4

Rewrite $\ln (\frac{x (x + 1)}{6}) = 0$ 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 $b^{y} = x$ .

$e^{0} = \frac{x (x + 1)}{6}$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $\frac{x (x + 1)}{6} = e^{0}$ .

$\frac{x (x + 1)}{6} = e^{0}$

Step 5.2

Multiply both sides of the equation by $6$ .

$6 \frac{x (x + 1)}{6} = 6 e^{0}$

Step 5.3

Simplify both sides of the equation.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Simplify $6 \frac{x (x + 1)}{6}$ .

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Step 5.3.1.1.1

Cancel the common factor of $6$ .

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Step 5.3.1.1.1.1

Cancel the common factor.

$6 \frac{x (x + 1)}{6} = 6 e^{0}$

Step 5.3.1.1.1.2

Rewrite the expression.

$x (x + 1) = 6 e^{0}$

Step 5.3.1.1.2

Apply the distributive property.

$x \cdot x + x \cdot 1 = 6 e^{0}$

Step 5.3.1.1.3

Simplify the expression.

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Step 5.3.1.1.3.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 = 6 e^{0}$

Step 5.3.1.1.3.2

Multiply $x$ by $1$ .

$x^{2} + x = 6 e^{0}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $6 e^{0}$ .

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Step 5.3.2.1.1

Anything raised to $0$ is $1$ .

$x^{2} + x = 6 \cdot 1$

Step 5.3.2.1.2

Multiply $6$ by $1$ .

$x^{2} + x = 6$

Step 5.4

Subtract $6$ from both sides of the equation.

$x^{2} + x - 6 = 0$

Step 5.5

Factor $x^{2} + x - 6$ using the AC method.

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Step 5.5.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 5.5.2

Write the factored form using these integers.

$(x - 2) (x + 3) = 0$

Step 5.6

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

$x - 2 = 0$

$x + 3 = 0$

Step 5.7

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 5.7.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 5.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5.8

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

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Step 5.8.1

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

$x + 3 = 0$

Step 5.8.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5.9

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$