What is the general solution for solvefor u,x=tanh(u) ?

The general solution for solvefor u,x=tanh(u) is u= 1/2 ln((x+1)/(1-x)),u=ln(-sqrt((x+1)/(1-x)))

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solvefor u,x=tanh(u)

Solution

solve for

u, x = tanh (u)

Solution

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

Solution steps

x = tanh (u)

Switch sides

tanh (u) = x

Rewrite using trig identities

tanh (u) = x

Use the Hyperbolic identity:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} = x

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} = x

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} = x : u = \frac{1}{2} ln (\frac{x + 1}{1 - x}) {- 1 < x < 1}, u = ln (- \frac{x + 1}{1 - x}) {No Solution}

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} = x

Multiply both sides by

e^{u} + e^{- u}

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} (e^{u} + e^{- u}) = x (e^{u} + e^{- u})

Simplify

e^{u} - e^{- u} = x (e^{u} + e^{- u})

Apply exponent rules

e^{u} - e^{- u} = x (e^{u} + e^{- u})

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- u} = (e^{u})^{- 1}

e^{u} - (e^{u})^{- 1} = x (e^{u} + (e^{u})^{- 1})

e^{u} - (e^{u})^{- 1} = x (e^{u} + (e^{u})^{- 1})

Rewrite the equation with

e^{u} = v

v - v^{- 1} = x (v + v^{- 1})

Solve

v - v^{- 1} = x (v + v^{- 1}) : v = \frac{x + 1}{1 - x}, v = - \frac{x + 1}{1 - x}

v - v^{- 1} = x (v + v^{- 1})

Refine

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

Multiply both sides by

v

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

Multiply both sides by

v

vv - \frac{1}{v} v = x (v + \frac{1}{v}) v

Simplify

vv - \frac{1}{v} v = x (v + \frac{1}{v}) v

Simplify

vv : v^{2}

vv

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

vv = v^{1 + 1}

= v^{1 + 1}

Add the numbers:

1 + 1 = 2

= v^{2}

Simplify

- \frac{1}{v} v : - 1

- \frac{1}{v} v

Multiply fractions:

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

= - \frac{1 \cdot v}{v}

Cancel the common factor:

v

= - 1

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

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

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

Expand

x (v + \frac{1}{v}) v : x v^{2} + x

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

= xv (v + \frac{1}{v})

Apply the distributive law:

a (b + c) = ab + a c

a = xv, b = v, c = \frac{1}{v}

= xvv + xv \frac{1}{v}

= xvv + \frac{1}{v} xv

Simplify

xvv + \frac{1}{v} xv : x v^{2} + x

xvv + \frac{1}{v} xv

xvv = x v^{2}

xvv

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

vv = v^{1 + 1}

= x v^{1 + 1}

Add the numbers:

1 + 1 = 2

= x v^{2}

\frac{1}{v} xv = x

\frac{1}{v} xv

Multiply fractions:

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

= \frac{1 \cdot xv}{v}

Cancel the common factor:

v

= 1 \cdot x

Multiply:

1 \cdot x = x

= x

= x v^{2} + x

= x v^{2} + x

v^{2} - 1 = x v^{2} + x

Move

1

to the right side

v^{2} - 1 = x v^{2} + x

Add

1

to both sides

v^{2} - 1 + 1 = x v^{2} + x + 1

Simplify

v^{2} = x v^{2} + x + 1

v^{2} = x v^{2} + x + 1

Solve

v^{2} = x v^{2} + x + 1 : v = \frac{x + 1}{1 - x}, v = - \frac{x + 1}{1 - x}

v^{2} = x v^{2} + x + 1

Move

x v^{2}

to the left side

v^{2} = x v^{2} + x + 1

Subtract

x v^{2}

from both sides

v^{2} - x v^{2} = x v^{2} + x + 1 - x v^{2}

Simplify

v^{2} - x v^{2} = x + 1

v^{2} - x v^{2} = x + 1

Factor

v^{2} - x v^{2} : v^{2} (1 - x)

v^{2} - x v^{2}

Factor out common term

v^{2}

= v^{2} (1 - x)

v^{2} (1 - x) = x + 1

Divide both sides by

1 - x

v^{2} (1 - x) = x + 1

Divide both sides by

1 - x

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

Simplify

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

Simplify

\frac{v ^{2} ( 1 - x )}{1 - x} : v^{2}

\frac{v ^{2} ( 1 - x )}{1 - x}

Cancel the common factor:

1 - x

= v^{2}

Simplify

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

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

Apply rule

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

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

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

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

v = \frac{x + 1}{1 - x}, v = - \frac{x + 1}{1 - x}

v = \frac{x + 1}{1 - x}, v = - \frac{x + 1}{1 - x}

v = \frac{x + 1}{1 - x}, v = - \frac{x + 1}{1 - x}

Substitute back

v = e^{u},

solve for

u

Solve

e^{u} = \frac{x + 1}{1 - x} : u = \frac{1}{2} ln (\frac{x + 1}{1 - x})

e^{u} = \frac{x + 1}{1 - x}

Apply exponent rules

e^{u} = \frac{x + 1}{1 - x}

Apply exponent rule:

a = a^{\frac{1}{2}}

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

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

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

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

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

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

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

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

Solve

e^{u} = - \frac{x + 1}{1 - x} : u = ln (- \frac{x + 1}{1 - x})

e^{u} = - \frac{x + 1}{1 - x}

Apply exponent rules

e^{u} = - \frac{x + 1}{1 - x}

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{u}) = ln (- \frac{x + 1}{1 - x})

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

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

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

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

Verify Solutions:

u = \frac{1}{2} ln (\frac{x + 1}{1 - x}) {- 1 < x < 1}, u = ln (- \frac{x + 1}{1 - x}) {No Solution}

Check the solutions by plugging them into

\frac{e ^{u} - e ^{- u}}{e ^{u} + e ^{- u}} = x

Remove the ones that don't agree with the equation.

Plug in

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

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}} = x

Simplify

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}} : x

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}

Factor

e^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})} : e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} (e^{l n (\frac{1 + x}{1 - x})} - 1)

e^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

e^{\frac{1}{2} l n (\frac{1 + x}{1 - x})} = e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})}

= e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})} - e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})}

Factor out common term

e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})}

= e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} (e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})} - 1)

Refine

= e^{- \frac{1}{2} l n (\frac{x + 1}{- x + 1})} (e^{l n (\frac{x + 1}{- x + 1})} - 1)

= \frac{e ^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} ( e ^{l n (\frac{1 + x}{1 - x})} - 1 )}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}

Factor

e^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})} : e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} (e^{l n (\frac{1 + x}{1 - x})} + 1)

e^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

e^{\frac{1}{2} l n (\frac{1 + x}{1 - x})} = e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})}

= e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})} + e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})}

Factor out common term

e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})}

= e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} (e^{2 \cdot \frac{1}{2} l n (\frac{1 + x}{1 - x})} + 1)

Refine

= e^{- \frac{1}{2} l n (\frac{x + 1}{- x + 1})} (e^{l n (\frac{x + 1}{- x + 1})} + 1)

= \frac{e ^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} ( e ^{l n (\frac{1 + x}{1 - x})} - 1 )}{e ^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})} ( e ^{l n (\frac{1 + x}{1 - x})} + 1 )}

Cancel the common factor:

e^{- \frac{1}{2} l n (\frac{1 + x}{1 - x})}

= \frac{e ^{l n (\frac{x + 1}{- x + 1})} - 1}{e ^{l n (\frac{x + 1}{- x + 1})} + 1}

Apply log rule:

a^{l o g_{a} (b)} = b

= \frac{e ^{l n (\frac{x + 1}{- x + 1})} - 1}{\frac{x + 1}{- x + 1} + 1}

Apply log rule:

a^{l o g_{a} (b)} = b

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

Simplify

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Multiply:

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

= x + 1 - x + 1

Remove parentheses:

(- a) = - a

= x + 1 - x + 1

Group like terms

= x - x + 1 + 1

Add similar elements:

x - x = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

Expand

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

x + 1 - (- x + 1)

- (- x + 1) : x - 1

- (- x + 1)

Distribute parentheses

= - (- x) - 1

Apply minus-plus rules

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

= x - 1

= x + 1 + x - 1

Simplify

x + 1 + x - 1 : 2 x

x + 1 + x - 1

Group like terms

= x + x + 1 - 1

Add similar elements:

x + x = 2 x

= 2 x + 1 - 1

1 - 1 = 0

= 2 x

= 2 x

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

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

- x + 1

= x

= x

x = x

Domain of

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}} : - 1 < x < 1

Domain definition

Find positive values for logs:

- 1 < x < 1

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}

lo g_{a} f (x) \Rightarrow f (x) > 0

Solve

\frac{x + 1}{1 - x} > 0 : - 1 < x < 1

\frac{x + 1}{1 - x} > 0

Identify the intervals

Find the signs of the factors of

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

Find the signs of

x + 1

x + 1 = 0 : x = - 1

x + 1 = 0

Move

1

to the right side

x + 1 = 0

Subtract

1

from both sides

x + 1 - 1 = 0 - 1

Simplify

x = - 1

x = - 1

x + 1 < 0 : x < - 1

x + 1 < 0

Move

1

to the right side

x + 1 < 0

Subtract

1

from both sides

x + 1 - 1 < 0 - 1

Simplify

x < - 1

x < - 1

x + 1 > 0 : x > - 1

x + 1 > 0

Move

1

to the right side

x + 1 > 0

Subtract

1

from both sides

x + 1 - 1 > 0 - 1

Simplify

x > - 1

x > - 1

Find the signs of

1 - x

1 - x = 0 : x = 1

1 - x = 0

Move

1

to the right side

1 - x = 0

Subtract

1

from both sides

1 - x - 1 = 0 - 1

Simplify

- x = - 1

- x = - 1

Divide both sides by

- 1

- x = - 1

Divide both sides by

- 1

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

Simplify

x = 1

x = 1

1 - x < 0 : x > 1

1 - x < 0

Move

1

to the right side

1 - x < 0

Subtract

1

from both sides

1 - x - 1 < 0 - 1

Simplify

- x < - 1

- x < - 1

Multiply both sides by

- 1

- x < - 1

Multiply both sides by -1 (reverse the inequality)

(- x) (- 1) > (- 1) (- 1)

Simplify

x > 1

x > 1

1 - x > 0 : x < 1

1 - x > 0

Move

1

to the right side

1 - x > 0

Subtract

1

from both sides

1 - x - 1 > 0 - 1

Simplify

- x > - 1

- x > - 1

Multiply both sides by

- 1

- x > - 1

Multiply both sides by -1 (reverse the inequality)

(- x) (- 1) < (- 1) (- 1)

Simplify

x < 1

x < 1

Find singularity points

Find the zeros of the denominator

1 - x :

No Solution

1 - x = 0

The sides are not equal

No Solution

Summarize in a table:

x + 1 1 - x \frac{x + 1}{1 - x} x < - 1 - + - x = - 1 0 + 0 - 1 < x < 1 + + + x = 1 + 0 Undefined x > 1 + - -

Identify the intervals that satisfy the required condition:

> 0

- 1 < x < 1

- 1 < x < 1

Find undefined (singularity) points:

x = 1

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}

Take the denominator(s) of

\frac{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} - e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}{e ^{\frac{1}{2} l n (\frac{x + 1}{1 - x})} + e ^{- \frac{1}{2} l n (\frac{x + 1}{1 - x})}}

and compare to zero

Solve

1 - x = 0 : x = 1

1 - x = 0

Move

1

to the right side

1 - x = 0

Subtract

1

from both sides

1 - x - 1 = 0 - 1

Simplify

- x = - 1

- x = - 1

Divide both sides by

- 1

- x = - 1

Divide both sides by

- 1

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

Simplify

x = 1

x = 1

The following points are undefined

x = 1

Combine real regions and undefined points for final function domain

- 1 < x < 1

- 1 < x < 1

Plug in

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

No Solution

\frac{e ^{l n (- \frac{x + 1}{1 - x})} - e ^{- l n (- \frac{x + 1}{1 - x})}}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}} = x

Simplify

\frac{e ^{l n (- \frac{x + 1}{1 - x})} - e ^{- l n (- \frac{x + 1}{1 - x})}}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}} : x

\frac{e ^{l n (- \frac{x + 1}{1 - x})} - e ^{- l n (- \frac{x + 1}{1 - x})}}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}}

Factor

e^{l n (- \frac{x + 1}{1 - x})} - e^{- l n (- \frac{x + 1}{1 - x})} : e^{- l n (- \frac{1 + x}{1 - x})} (e^{2 l n (- \frac{1 + x}{1 - x})} - 1)

e^{l n (- \frac{x + 1}{1 - x})} - e^{- l n (- \frac{x + 1}{1 - x})}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

e^{l n (- \frac{1 + x}{1 - x})} = e^{- l n (- \frac{1 + x}{1 - x})} e^{2 l n (- \frac{1 + x}{1 - x})}

= e^{- l n (- \frac{1 + x}{1 - x})} e^{2 l n (- \frac{1 + x}{1 - x})} - e^{- l n (- \frac{1 + x}{1 - x})}

Factor out common term

e^{- l n (- \frac{1 + x}{1 - x})}

= e^{- l n (- \frac{1 + x}{1 - x})} (e^{2 l n (- \frac{1 + x}{1 - x})} - 1)

= \frac{e ^{- l n (- \frac{1 + x}{1 - x})} ( e ^{2 l n (- \frac{1 + x}{1 - x})} - 1 )}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}}

Factor

e^{l n (- \frac{x + 1}{1 - x})} + e^{- l n (- \frac{x + 1}{1 - x})} : e^{- l n (- \frac{1 + x}{1 - x})} (e^{2 l n (- \frac{1 + x}{1 - x})} + 1)

e^{l n (- \frac{x + 1}{1 - x})} + e^{- l n (- \frac{x + 1}{1 - x})}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

e^{l n (- \frac{1 + x}{1 - x})} = e^{- l n (- \frac{1 + x}{1 - x})} e^{2 l n (- \frac{1 + x}{1 - x})}

= e^{- l n (- \frac{1 + x}{1 - x})} e^{2 l n (- \frac{1 + x}{1 - x})} + e^{- l n (- \frac{1 + x}{1 - x})}

Factor out common term

e^{- l n (- \frac{1 + x}{1 - x})}

= e^{- l n (- \frac{1 + x}{1 - x})} (e^{2 l n (- \frac{1 + x}{1 - x})} + 1)

= \frac{e ^{- l n (- \frac{1 + x}{1 - x})} ( e ^{2 l n (- \frac{1 + x}{1 - x})} - 1 )}{e ^{- l n (- \frac{1 + x}{1 - x})} ( e ^{2 l n (- \frac{1 + x}{1 - x})} + 1 )}

Cancel the common factor:

e^{- l n (- \frac{1 + x}{1 - x})}

= \frac{e ^{2 l n (- \frac{x + 1}{- x + 1})} - 1}{e ^{2 l n (- \frac{x + 1}{- x + 1})} + 1}

e^{2 l n (- \frac{x + 1}{- x + 1})} = \frac{x + 1}{- x + 1}

e^{2 l n (- \frac{x + 1}{- x + 1})}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (- \frac{x + 1}{- x + 1})})^{2}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (- \frac{x + 1}{- x + 1})} = - \frac{x + 1}{- x + 1}

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

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

= \frac{e ^{2 l n (- \frac{x + 1}{- x + 1})} - 1}{\frac{x + 1}{- x + 1} + 1}

e^{2 l n (- \frac{x + 1}{- x + 1})} = \frac{x + 1}{- x + 1}

e^{2 l n (- \frac{x + 1}{- x + 1})}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (- \frac{x + 1}{- x + 1})})^{2}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (- \frac{x + 1}{- x + 1})} = - \frac{x + 1}{- x + 1}

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

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

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

Simplify

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Multiply:

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

= x + 1 - x + 1

Remove parentheses:

(- a) = - a

= x + 1 - x + 1

Group like terms

= x - x + 1 + 1

Add similar elements:

x - x = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

Expand

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

x + 1 - (- x + 1)

- (- x + 1) : x - 1

- (- x + 1)

Distribute parentheses

= - (- x) - 1

Apply minus-plus rules

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

= x - 1

= x + 1 + x - 1

Simplify

x + 1 + x - 1 : 2 x

x + 1 + x - 1

Group like terms

= x + x + 1 - 1

Add similar elements:

x + x = 2 x

= 2 x + 1 - 1

1 - 1 = 0

= 2 x

= 2 x

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

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

- x + 1

= x

= x

x = x

Domain of

\frac{e ^{l n (- \frac{x + 1}{1 - x})} - e ^{- l n (- \frac{x + 1}{1 - x})}}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}} :

Undefined

Domain definition

Find positive values for logs:

No Solution

\frac{e ^{l n (- \frac{x + 1}{1 - x})} - e ^{- l n (- \frac{x + 1}{1 - x})}}{e ^{l n (- \frac{x + 1}{1 - x})} + e ^{- l n (- \frac{x + 1}{1 - x})}}

lo g_{a} f (x) \Rightarrow f (x) > 0

Solve

- \frac{x + 1}{1 - x} > 0 :

No Solution

- \frac{x + 1}{1 - x} > 0

Multiply both sides by

- 1

(reverse the inequality)

(- \frac{x + 1}{1 - x}) (- 1) < 0 \cdot (- 1)

Simplify

\frac{x + 1}{1 - x} < 0

If n is even,

for all

u

No Solution for x \in R

Solve

- \frac{x + 1}{1 - x} > 0 :

No Solution

- \frac{x + 1}{1 - x} > 0

Multiply both sides by

- 1

(reverse the inequality)

(- \frac{x + 1}{1 - x}) (- 1) < 0 \cdot (- 1)

Simplify

\frac{x + 1}{1 - x} < 0

If n is even,

for all

u

No Solution for x \in R

Combine the intervals

No Solution

Undefined

No Solution

The solutions are

u = \frac{1}{2} ln (\frac{x + 1}{1 - x}) {- 1 < x < 1}, u = ln (- \frac{x + 1}{1 - x}) {No Solution}

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

Popular Examples

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tan^2(x)=3-2tan(x)

tan^{2} (x) = 3 - 2 tan (x)

tan(3x)+tan(5x)=0

tan (3 x) + tan (5 x) = 0

sin(x)= 1/2 sqrt(3)

sin (x) = \frac{1}{2} 3

Frequently Asked Questions (FAQ)

What is the general solution for solvefor u,x=tanh(u) ?
The general solution for solvefor u,x=tanh(u) is u= 1/2 ln((x+1)/(1-x)),u=ln(-sqrt((x+1)/(1-x)))