Algebra Examples

$\ln (\frac{1 + x y}{1 - x y}) - 2 x = c$

Step 1

Add $2 x$ to both sides of the equation.

$\ln (\frac{1 + x y}{1 - x y}) = c + 2 x$

Step 2

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

$e^{\ln (\frac{1 + x y}{1 - x y})} = e^{c + 2 x}$

Step 3

Rewrite $\ln (\frac{1 + x y}{1 - x y}) = c + 2 x$ 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 $b^{y} = x$ .

$e^{c + 2 x} = \frac{1 + x y}{1 - x y}$

Step 4

Solve for $y$ .

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Step 4.1

Rewrite the equation as $\frac{1 + x y}{1 - x y} = e^{c + 2 x}$ .

$\frac{1 + x y}{1 - x y} = e^{c + 2 x}$

Step 4.2

Find the LCD of the terms in the equation.

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Step 4.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1 - x y, 1$

Step 4.2.2

Remove parentheses.

$1 - x y, 1$

Step 4.2.3

The LCM of one and any expression is the expression.

$1 - x y$

Step 4.3

Multiply each term in $\frac{1 + x y}{1 - x y} = e^{c + 2 x}$ by $1 - x y$ to eliminate the fractions.

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Step 4.3.1

Multiply each term in $\frac{1 + x y}{1 - x y} = e^{c + 2 x}$ by $1 - x y$ .

$\frac{1 + x y}{1 - x y} (1 - x y) = e^{c + 2 x} (1 - x y)$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $1 - x y$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{1 + x y}{1 - x y} (1 - x y) = e^{c + 2 x} (1 - x y)$

Step 4.3.2.1.2

Rewrite the expression.

$1 + x y = e^{c + 2 x} (1 - x y)$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Apply the distributive property.

$1 + x y = e^{c + 2 x} \cdot 1 + e^{c + 2 x} (- x y)$

Step 4.3.3.2

Simplify the expression.

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Step 4.3.3.2.1

Multiply $e^{c + 2 x}$ by $1$ .

$1 + x y = e^{c + 2 x} + e^{c + 2 x} (- x y)$

Step 4.3.3.2.2

Rewrite using the commutative property of multiplication.

$1 + x y = e^{c + 2 x} - e^{c + 2 x} (x y)$

Step 4.3.3.2.3

Reorder factors in $e^{c + 2 x} - e^{c + 2 x} (x y)$ .

$1 + x y = e^{c + 2 x} - x y e^{c + 2 x}$

Step 4.4

Solve the equation.

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Step 4.4.1

Add $x y e^{c + 2 x}$ to both sides of the equation.

$1 + x y + x y e^{c + 2 x} = e^{c + 2 x}$

Step 4.4.2

Subtract $1$ from both sides of the equation.

$x y + x y e^{c + 2 x} = e^{c + 2 x} - 1$

Step 4.4.3

Factor $x y$ out of $x y + x y e^{c + 2 x}$ .

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Step 4.4.3.1

Factor $x y$ out of $x y$ .

$x y (1) + x y e^{c + 2 x} = e^{c + 2 x} - 1$

Step 4.4.3.2

Factor $x y$ out of $x y e^{c + 2 x}$ .

$x y (1) + x y (e^{c + 2 x}) = e^{c + 2 x} - 1$

Step 4.4.3.3

Factor $x y$ out of $x y (1) + x y (e^{c + 2 x})$ .

$x y (1 + e^{c + 2 x}) = e^{c + 2 x} - 1$

Step 4.4.4

Divide each term in $x y (1 + e^{c + 2 x}) = e^{c + 2 x} - 1$ by $x (1 + e^{c + 2 x})$ and simplify.

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Step 4.4.4.1

Divide each term in $x y (1 + e^{c + 2 x}) = e^{c + 2 x} - 1$ by $x (1 + e^{c + 2 x})$ .

$\frac{x y (1 + e^{c + 2 x})}{x (1 + e^{c + 2 x})} = \frac{e^{c + 2 x}}{x (1 + e^{c + 2 x})} + \frac{- 1}{x (1 + e^{c + 2 x})}$

Step 4.4.4.2

Simplify the left side.

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Step 4.4.4.2.1

Cancel the common factor of $x$ .

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Step 4.4.4.2.1.1

Cancel the common factor.

$\frac{x y (1 + e^{c + 2 x})}{x (1 + e^{c + 2 x})} = \frac{e^{c + 2 x}}{x (1 + e^{c + 2 x})} + \frac{- 1}{x (1 + e^{c + 2 x})}$

Step 4.4.4.2.1.2

Rewrite the expression.

$\frac{y (1 + e^{c + 2 x})}{1 + e^{c + 2 x}} = \frac{e^{c + 2 x}}{x (1 + e^{c + 2 x})} + \frac{- 1}{x (1 + e^{c + 2 x})}$

Step 4.4.4.2.2

Cancel the common factor of $1 + e^{c + 2 x}$ .

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Step 4.4.4.2.2.1

Cancel the common factor.

$\frac{y (1 + e^{c + 2 x})}{1 + e^{c + 2 x}} = \frac{e^{c + 2 x}}{x (1 + e^{c + 2 x})} + \frac{- 1}{x (1 + e^{c + 2 x})}$

Step 4.4.4.2.2.2

Divide $y$ by $1$ .

$y = \frac{e^{c + 2 x}}{x (1 + e^{c + 2 x})} + \frac{- 1}{x (1 + e^{c + 2 x})}$

Step 4.4.4.3

Simplify the right side.

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Step 4.4.4.3.1

Combine the numerators over the common denominator.

$y = \frac{e^{c + 2 x} - 1}{x (1 + e^{c + 2 x})}$