What is the general solution for tanh(x)= 3/5 ?

The general solution for tanh(x)= 3/5 is x=ln(2)

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Popular Trigonometry >

tanh(x)= 3/5

Solution

tanh (x) = \frac{3}{5}

Solution

x = ln (2)

Degrees

x = 39.71440 \dots^{\circ}

Solution steps

tanh (x) = \frac{3}{5}

Rewrite using trig identities

tanh (x) = \frac{3}{5}

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{3}{5}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5} : x = ln (2)

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

(e^{x} - e^{- x}) \cdot 5 = (e^{x} + e^{- x}) \cdot 3

Apply exponent rules

(e^{x} - e^{- x}) \cdot 5 = (e^{x} + e^{- x}) \cdot 3

Apply exponent rule:

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

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

(e^{x} - (e^{x})^{- 1}) \cdot 5 = (e^{x} + (e^{x})^{- 1}) \cdot 3

(e^{x} - (e^{x})^{- 1}) \cdot 5 = (e^{x} + (e^{x})^{- 1}) \cdot 3

Rewrite the equation with

e^{x} = u

(u - (u)^{- 1}) \cdot 5 = (u + (u)^{- 1}) \cdot 3

Solve

(u - u^{- 1}) \cdot 5 = (u + u^{- 1}) \cdot 3 : u = 2, u = - 2

(u - u^{- 1}) \cdot 5 = (u + u^{- 1}) \cdot 3

Refine

(u - \frac{1}{u}) \cdot 5 = (u + \frac{1}{u}) \cdot 3

Simplify

(u - \frac{1}{u}) \cdot 5 = (u + \frac{1}{u}) \cdot 3

Simplify

(u - \frac{1}{u}) \cdot 5 : 5 (u - \frac{1}{u})

(u - \frac{1}{u}) \cdot 5

Apply the commutative law:

(u - \frac{1}{u}) \cdot 5 = 5 (u - \frac{1}{u})

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

Simplify

(u + \frac{1}{u}) \cdot 3 : 3 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 3

Apply the commutative law:

(u + \frac{1}{u}) \cdot 3 = 3 (u + \frac{1}{u})

3 (u + \frac{1}{u})

5 (u - \frac{1}{u}) = 3 (u + \frac{1}{u})

5 (u - \frac{1}{u}) = 3 (u + \frac{1}{u})

Expand

5 (u - \frac{1}{u}) : 5 u - \frac{5}{u}

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

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = u, c = \frac{1}{u}

= 5 u - 5 \cdot \frac{1}{u}

5 \cdot \frac{1}{u} = \frac{5}{u}

5 \cdot \frac{1}{u}

Multiply fractions:

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

= \frac{1 \cdot 5}{u}

Multiply the numbers:

1 \cdot 5 = 5

= \frac{5}{u}

= 5 u - \frac{5}{u}

Expand

3 (u + \frac{1}{u}) : 3 u + \frac{3}{u}

3 (u + \frac{1}{u})

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = u, c = \frac{1}{u}

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

3 \cdot \frac{1}{u} = \frac{3}{u}

3 \cdot \frac{1}{u}

Multiply fractions:

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

= \frac{1 \cdot 3}{u}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{u}

= 3 u + \frac{3}{u}

5 u - \frac{5}{u} = 3 u + \frac{3}{u}

Multiply both sides by

u

5 u - \frac{5}{u} = 3 u + \frac{3}{u}

Multiply both sides by

u

5 uu - \frac{5}{u} u = 3 uu + \frac{3}{u} u

Simplify

5 uu - \frac{5}{u} u = 3 uu + \frac{3}{u} u

Simplify

5 uu : 5 u^{2}

5 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 5 u^{2}

Simplify

- \frac{5}{u} u : - 5

- \frac{5}{u} u

Multiply fractions:

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

= - \frac{5 u}{u}

Cancel the common factor:

u

= - 5

Simplify

3 uu : 3 u^{2}

3 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 3 u^{2}

Simplify

\frac{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

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

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

5 u^{2} - 5 = 3 u^{2} + 3

5 u^{2} - 5 = 3 u^{2} + 3

5 u^{2} - 5 = 3 u^{2} + 3

Solve

5 u^{2} - 5 = 3 u^{2} + 3 : u = 2, u = - 2

5 u^{2} - 5 = 3 u^{2} + 3

Move

5

to the right side

5 u^{2} - 5 = 3 u^{2} + 3

Add

5

to both sides

5 u^{2} - 5 + 5 = 3 u^{2} + 3 + 5

Simplify

5 u^{2} = 3 u^{2} + 8

5 u^{2} = 3 u^{2} + 8

Move

3 u^{2}

to the left side

5 u^{2} = 3 u^{2} + 8

Subtract

3 u^{2}

from both sides

5 u^{2} - 3 u^{2} = 3 u^{2} + 8 - 3 u^{2}

Simplify

2 u^{2} = 8

2 u^{2} = 8

Divide both sides by

2

2 u^{2} = 8

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{8}{2}

Simplify

u^{2} = 4

u^{2} = 4

For

x^{2} = f (a)

the solutions are

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

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

- 4 = - 2

- 4

Factor the number:

4 = 2^{2}

= - 2^{2}

Apply radical rule:

a^{2} = a, a \geq 0

2^{2} = - 2

= - 2

u = 2, u = - 2

u = 2, u = - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u - u^{- 1}) 5

and compare to zero

u = 0

Take the denominator(s) of

(u + u^{- 1}) 3

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2, u = - 2

u = 2, u = - 2

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 : x = ln (2)

e^{x} = 2

Apply exponent rules

e^{x} = 2

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2)

x = ln (2)

Solve

e^{x} = - 2 :

No Solution for

x \in R

e^{x} = - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (2)

Verify Solutions:

x = ln (2)

True

Check the solutions by plugging them into

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

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

Plug in

x = ln (2) :

True

\frac{e ^{l n (2)} - e ^{- l n (2)}}{e ^{l n (2)} + e ^{- l n (2)}} = \frac{3}{5}

\frac{e ^{l n (2)} - e ^{- l n (2)}}{e ^{l n (2)} + e ^{- l n (2)}} = \frac{3}{5}

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

e^{l n (2)} = 2

e^{l n (2)}

Apply log rule:

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

= 2

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

e^{- l n (2)}

Apply exponent rule:

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

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

Apply log rule:

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

e^{l n (2)} = 2

= 2^{- 1}

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

e^{l n (2)} = 2

e^{l n (2)}

Apply log rule:

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

= 2

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

e^{- l n (2)}

Apply exponent rule:

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

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

Apply log rule:

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

e^{l n (2)} = 2

= 2^{- 1}

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

Simplify

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

Join

2 + \frac{1}{2} : \frac{5}{2}

2 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 + 1

Add the numbers:

4 + 1 = 5

= 5

= \frac{5}{2}

= \frac{2 - \frac{1}{2}}{\frac{5}{2}}

Join

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

2 - \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

2 \cdot 2 - 1 = 3

2 \cdot 2 - 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 - 1

Subtract the numbers:

4 - 1 = 3

= 3

= \frac{3}{2}

= \frac{\frac{3}{2}}{\frac{5}{2}}

Divide fractions:

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

= \frac{3 \cdot 2}{2 \cdot 5}

Cancel the common factor:

2

= \frac{3}{5}

= \frac{3}{5}

\frac{3}{5} = \frac{3}{5}

True

The solution is

x = ln (2)

x = ln (2)

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Frequently Asked Questions (FAQ)

What is the general solution for tanh(x)= 3/5 ?
The general solution for tanh(x)= 3/5 is x=ln(2)