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

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

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tanh(x+1)+1=2

tanh (x + 1) + 1 = 2

No Solution for x \in R

Solution steps

tanh (x + 1) + 1 = 2

Rewrite using trig identities

tanh (x + 1) + 1 = 2

Use the Hyperbolic identity:

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

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

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

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

No Solution for

x \in R

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

Multiply both sides by

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

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

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

No Solution for

u \in R

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

Refine

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

Expand

2 (e u + \frac{1}{e u}) : 2 e u + \frac{2}{e u}

2 (e u + \frac{1}{e u})

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = e u, c = \frac{1}{e u}

= 2 e u + 2 \cdot \frac{1}{e u}

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

2 \cdot \frac{1}{e u}

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{e u}

= 2 e u + \frac{2}{e u}

2 e u = 2 e u + \frac{2}{e u}

Subtract

2 e u

from both sides

2 e u - 2 e u = 2 e u + \frac{2}{e u} - 2 e u

Simplify

0 = \frac{2}{e u}

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

0 = 2

The sides are not equal

No Solution for u \in R

No Solution for x \in R

No Solution for x \in R

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