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

The general solution for ((1+tanh(x))/(1-tanh(x)))^2=2 is x= 1/4 ln(2)

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((1+tanh(x))/(1-tanh(x)))^2=2

Solution

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

Solution

x = \frac{1}{4} ln (2)

Degrees

x = 9.92860 \dots^{\circ}

Solution steps

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

Rewrite using trig identities

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

Use the Hyperbolic identity:

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

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

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

(\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}})^{2} = 2 : x = \frac{1}{4} ln (2)

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

Refine

\frac{4 u ^{4}}{( u ^{2} + 1 - ( u ^{2} - 1 ) ) ^{2}} = 2

Simplify

\frac{4 u ^{4}}{( u ^{2} + 1 - ( u ^{2} - 1 ) ) ^{2}} : u^{4}

\frac{4 u ^{4}}{( u ^{2} + 1 - ( u ^{2} - 1 ) ) ^{2}}

Expand

u^{2} + 1 - (u^{2} - 1) : 2

u^{2} + 1 - (u^{2} - 1)

- (u^{2} - 1) : - u^{2} + 1

- (u^{2} - 1)

Distribute parentheses

= - (u^{2}) - (- 1)

Apply minus-plus rules

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

= - u^{2} + 1

= u^{2} + 1 - u^{2} + 1

Simplify

u^{2} + 1 - u^{2} + 1 : 2

u^{2} + 1 - u^{2} + 1

Group like terms

= u^{2} - u^{2} + 1 + 1

Add similar elements:

u^{2} - u^{2} = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

= \frac{4 u ^{4}}{2 ^{2}}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

= \frac{2 ^{2} u ^{4}}{2 ^{2}}

Cancel the common factor:

2^{2}

= u^{4}

u^{4} = 2

Solve

u^{4} = 2

For

x^{n} = f (a)

, n is even, the solutions are

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

Substitute back

u = e^{x},

solve for

x

Solve

Apply exponent rules

Apply exponent rule:

e^{x} = 2^{\frac{1}{4}}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2^{\frac{1}{4}})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2^{\frac{1}{4}})

Apply log rule:

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

ln (2^{\frac{1}{4}}) = \frac{1}{4} ln (2)

x = \frac{1}{4} ln (2)

x = \frac{1}{4} ln (2)

Solve

No Solution for

x \in R

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = \frac{1}{4} ln (2)

x = \frac{1}{4} ln (2)

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(1+tanh(x))/(1-tanh(x))=2

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

Frequently Asked Questions (FAQ)

What is the general solution for ((1+tanh(x))/(1-tanh(x)))^2=2 ?
The general solution for ((1+tanh(x))/(1-tanh(x)))^2=2 is x= 1/4 ln(2)