What is the general solution for 2arctanh(((x-2))/((x+1)))=ln(2) ?

The general solution for 2arctanh(((x-2))/((x+1)))=ln(2) is No Solution for x\in\mathbb{R}

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2arctanh(((x-2))/((x+1)))=ln(2)

Solution

2 arctanh (\frac{( x - 2 )}{( x + 1 )}) = ln (2)

Solution

No Solution for x \in R

Solution steps

2 arctanh (\frac{( x - 2 )}{( x + 1 )}) = ln (2)

Solve by substitution

2 arctanh (\frac{x - 2}{x + 1}) = ln (2)

Let:

arctanh (\frac{x - 2}{x + 1}) = u

2 u = ln (2)

2 u = ln (2) : u = \frac{ln ( 2 )}{2}

2 u = ln (2)

Divide both sides by

2

2 u = ln (2)

Divide both sides by

2

\frac{2 u}{2} = \frac{ln ( 2 )}{2}

Simplify

u = \frac{ln ( 2 )}{2}

u = \frac{ln ( 2 )}{2}

Substitute back

u = arctanh (\frac{x - 2}{x + 1})

arctanh (\frac{x - 2}{x + 1}) = \frac{ln ( 2 )}{2}

arctanh (\frac{x - 2}{x + 1}) = \frac{ln ( 2 )}{2}

Subtract

ln (2)

from both sides

2 arctanh (\frac{x - 2}{x + 1}) - ln (2) = 0

Solve by substitution

2 arctanh (\frac{x - 2}{x + 1}) - ln (2) = 0

Let:

arctanh (\frac{x - 2}{x + 1}) = u

2 u - ln (2) = 0

2 u - ln (2) = 0 : u = \frac{ln ( 2 )}{2}

2 u - ln (2) = 0

Move

ln (2)

to the right side

2 u - ln (2) = 0

Add

ln (2)

to both sides

2 u - ln (2) + ln (2) = 0 + ln (2)

Simplify

2 u = ln (2)

2 u = ln (2)

Divide both sides by

2

2 u = ln (2)

Divide both sides by

2

\frac{2 u}{2} = \frac{ln ( 2 )}{2}

Simplify

u = \frac{ln ( 2 )}{2}

u = \frac{ln ( 2 )}{2}

Substitute back

u = arctanh (\frac{x - 2}{x + 1})

arctanh (\frac{x - 2}{x + 1}) = \frac{ln ( 2 )}{2}

arctanh (\frac{x - 2}{x + 1}) = \frac{ln ( 2 )}{2}

Rewrite using trig identities

- ln (2) + 2 arctanh (\frac{- 2 + x}{1 + x})

Use the Hyperbolic identity:

arctanh (x) = \frac{ln ( \frac{1 + x}{1 - x} )}{2}

= - ln (2) + 2 \cdot \frac{ln ( \frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}} )}{2}

2 \cdot \frac{ln ( \frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}} )}{2} = ln (\frac{2 x - 1}{3})

2 \cdot \frac{ln ( \frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}} )}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{ln ( \frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}} ) \cdot 2}{2}

Cancel the common factor:

2

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

\frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}} = \frac{2 x - 1}{3}

\frac{1 + \frac{- 2 + x}{1 + x}}{1 - \frac{- 2 + x}{1 + x}}

Join

1 - \frac{- 2 + x}{1 + x} : \frac{3}{1 + x}

1 - \frac{- 2 + x}{1 + x}

Convert element to fraction:

1 = \frac{1 ( 1 + x )}{1 + x}

= \frac{1 \cdot ( 1 + x )}{1 + x} - \frac{- 2 + x}{1 + x}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot ( 1 + x ) - ( - 2 + x )}{1 + x}

1 \cdot (1 + x) = 1 + x

1 \cdot (1 + x)

Multiply:

1 \cdot (1 + x) = (1 + x)

= (1 + x)

Remove parentheses:

(a) = a

= 1 + x

= \frac{1 + x - ( x - 2 )}{1 + x}

Expand

1 + x - (- 2 + x) : 3

1 + x - (- 2 + x)

- (- 2 + x) : 2 - x

- (- 2 + x)

Distribute parentheses

= - (- 2) - (x)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 2 - x

= 1 + x + 2 - x

Simplify

1 + x + 2 - x : 3

1 + x + 2 - x

Group like terms

= x - x + 1 + 2

Add similar elements:

x - x = 0

= 1 + 2

Add the numbers:

1 + 2 = 3

= 3

= 3

= \frac{3}{1 + x}

= \frac{1 + \frac{x - 2}{x + 1}}{\frac{3}{1 + x}}

Join

1 + \frac{- 2 + x}{1 + x} : \frac{2 x - 1}{1 + x}

1 + \frac{- 2 + x}{1 + x}

Convert element to fraction:

1 = \frac{1 ( 1 + x )}{1 + x}

= \frac{1 \cdot ( 1 + x )}{1 + x} + \frac{- 2 + x}{1 + x}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot ( 1 + x ) - 2 + x}{1 + x}

1 \cdot (1 + x) - 2 + x = 2 x - 1

1 \cdot (1 + x) - 2 + x

1 \cdot (1 + x) = 1 + x

1 \cdot (1 + x)

Multiply:

1 \cdot (1 + x) = (1 + x)

= (1 + x)

Remove parentheses:

(a) = a

= 1 + x

= 1 + x - 2 + x

Group like terms

= x + x + 1 - 2

Add similar elements:

x + x = 2 x

= 2 x + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= 2 x - 1

= \frac{2 x - 1}{1 + x}

= \frac{\frac{2 x - 1}{1 + x}}{\frac{3}{1 + x}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( 2 x - 1 ) ( 1 + x )}{( 1 + x ) \cdot 3}

Cancel the common factor:

1 + x

= \frac{2 x - 1}{3}

= ln (\frac{2 x - 1}{3})

= - ln (2) + ln (\frac{2 x - 1}{3})

- ln (2) + ln (\frac{- 1 + 2 x}{3}) = 0

Move

ln (2)

to the right side

- ln (2) + ln (\frac{- 1 + 2 x}{3}) = 0

Add

ln (2)

to both sides

- ln (2) + ln (\frac{- 1 + 2 x}{3}) + ln (2) = 0 + ln (2)

Simplify

ln (\frac{- 1 + 2 x}{3}) = ln (2)

ln (\frac{- 1 + 2 x}{3}) = ln (2)

Apply trig inverse properties

ln (\frac{- 1 + 2 x}{3}) = ln (2)

General solutions for

ln (\frac{- 1 + 2 x}{3}) = ln (2)

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2arctanh(((x-2))/((x+1)))=ln(2) ?
The general solution for 2arctanh(((x-2))/((x+1)))=ln(2) is No Solution for x\in\mathbb{R}