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

The general solution for tanh(x)= 1/2 is x= 1/2 ln(3)

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tanh(x)= 1/2

Solution

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

Solution

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

Degrees

x = 31.47292 \dots^{\circ}

Solution steps

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

Rewrite using trig identities

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

Use the Hyperbolic identity:

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

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

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

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

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

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

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

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

(u - u^{- 1}) \cdot 2 = u + u^{- 1} : u = 3, u = - 3

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

Refine

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

Simplify

(u - \frac{1}{u}) \cdot 2 : 2 (u - \frac{1}{u})

(u - \frac{1}{u}) \cdot 2

Apply the commutative law:

(u - \frac{1}{u}) \cdot 2 = 2 (u - \frac{1}{u})

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

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

Cancel the common factor:

u

= 1

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

= 2 uu - 2 u \frac{1}{u}

= 2 uu - 2 \cdot \frac{1}{u} u

Simplify

2 uu - 2 \cdot \frac{1}{u} u : 2 u^{2} - 2

2 uu - 2 \cdot \frac{1}{u} u

2 uu = 2 u^{2}

2 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

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

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

Multiply fractions:

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

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

Cancel the common factor:

u

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

= 2 u^{2} - 2

= 2 u^{2} - 2

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

Move

2

to the right side

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

Add

2

to both sides

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

Simplify

2 u^{2} = u^{2} + 3

2 u^{2} = u^{2} + 3

Solve

2 u^{2} = u^{2} + 3 : u = 3, u = - 3

2 u^{2} = u^{2} + 3

Move

u^{2}

to the left side

2 u^{2} = u^{2} + 3

Subtract

u^{2}

from both sides

2 u^{2} - u^{2} = u^{2} + 3 - u^{2}

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

u = 3, u = - 3

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u - u^{- 1}) 2

and compare to zero

u = 0

Take the denominator(s) of

u + u^{- 1}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 3, u = - 3

u = 3, u = - 3

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 3 : x = \frac{1}{2} ln (3)

e^{x} = 3

Apply exponent rules

e^{x} = 3

Apply exponent rule:

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

3 = 3^{\frac{1}{2}}

e^{x} = 3^{\frac{1}{2}}

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

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

Apply log rule:

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

ln (3^{\frac{1}{2}}) = \frac{1}{2} ln (3)

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

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

Solve

e^{x} = - 3 :

No Solution for

x \in R

e^{x} = - 3

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

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

Verify Solutions:

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

True

Check the solutions by plugging them into

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

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

Plug in

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

True

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}} = \frac{1}{2}

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}} = \frac{1}{2}

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}}

e^{\frac{1}{2} l n (3)} = 3

e^{\frac{1}{2} l n (3)}

Apply exponent rule:

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

= e^{l n (3)}

Apply log rule:

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

e^{l n (3)} = 3

= 3

e^{- \frac{1}{2} l n (3)} = 3^{- \frac{1}{2}}

e^{- \frac{1}{2} l n (3)}

Apply exponent rule:

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

= (e^{l n (3)})^{- \frac{1}{2}}

Apply log rule:

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

e^{l n (3)} = 3

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

= \frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{3 + 3 ^{- \frac{1}{2}}}

e^{\frac{1}{2} l n (3)} = 3

e^{\frac{1}{2} l n (3)}

Apply exponent rule:

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

= e^{l n (3)}

Apply log rule:

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

e^{l n (3)} = 3

= 3

e^{- \frac{1}{2} l n (3)} = 3^{- \frac{1}{2}}

e^{- \frac{1}{2} l n (3)}

Apply exponent rule:

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

= (e^{l n (3)})^{- \frac{1}{2}}

Apply log rule:

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

e^{l n (3)} = 3

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

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

Simplify

\frac{3 - 3 ^{- \frac{1}{2}}}{3 + 3 ^{- \frac{1}{2}}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

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

Join

3 + \frac{1}{3} : \frac{4}{3}

3 + \frac{1}{3}

Convert element to fraction:

3 = \frac{3 3}{3}

= \frac{3 3}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{3 3 + 1}{3}

33 + 1 = 4

33 + 1

Apply radical rule:

a a = a

33 = 3

= 3 + 1

Add the numbers:

3 + 1 = 4

= 4

= \frac{4}{3}

= \frac{3 - \frac{1}{3}}{\frac{4}{3}}

Join

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

3 - \frac{1}{3}

Convert element to fraction:

3 = \frac{3 3}{3}

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

Since the denominators are equal, combine the fractions:

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

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

33 - 1 = 2

33 - 1

Apply radical rule:

a a = a

33 = 3

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= \frac{2}{3}

= \frac{\frac{2}{3}}{\frac{4}{3}}

Divide fractions:

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

= \frac{2 3}{3 \cdot 4}

Cancel the common factor:

3

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

= \frac{1}{2}

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

True

The solution is

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

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

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Frequently Asked Questions (FAQ)

What is the general solution for tanh(x)= 1/2 ?
The general solution for tanh(x)= 1/2 is x= 1/2 ln(3)