What is the general solution for 4tanh(x)-1/(cosh(x))=1 ?

The general solution for 4tanh(x)-1/(cosh(x))=1 is x=ln(5/3)

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4tanh(x)-1/(cosh(x))=1

Solution

4 tanh (x) - \frac{1}{cosh ( x )} = 1

Solution

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

Degrees

x = 29.26815 \dots^{\circ}

Solution steps

4 tanh (x) - \frac{1}{cosh ( x )} = 1

Rewrite using trig identities

4 tanh (x) - \frac{1}{cosh ( x )} = 1

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

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

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

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

4 \cdot \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} - \frac{1}{\frac{e ^{x} + e ^{- x}}{2}} = 1 : x = ln (\frac{5}{3})

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

Multiply both sides by

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

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

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

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

Multiply both sides by

2 u

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

Multiply both sides by

2 u

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

Simplify

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

Simplify

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

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

Multiply the numbers:

2 \cdot 2 = 4

= 4 u (u - \frac{1}{u})

Simplify

- 1 \cdot 2 u : - 2 u

- 1 \cdot 2 u

Multiply the numbers:

1 \cdot 2 = 2

= - 2 u

Simplify

\frac{u ^{2} + 1}{2 u} \cdot 2 u : u^{2} + 1

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Cancel the common factor:

u

= u^{2} + 1

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

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

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

Expand

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 4 u, b = u, c = \frac{1}{u}

= 4 uu - 4 u \frac{1}{u}

= 4 uu - 4 \cdot \frac{1}{u} u

Simplify

4 uu - 4 \cdot \frac{1}{u} u : 4 u^{2} - 4

4 uu - 4 \cdot \frac{1}{u} u

4 uu = 4 u^{2}

4 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 u^{2}

4 \cdot \frac{1}{u} u = 4

4 \cdot \frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot 4 u}{u}

Cancel the common factor:

u

= 1 \cdot 4

Multiply the numbers:

1 \cdot 4 = 4

= 4

= 4 u^{2} - 4

= 4 u^{2} - 4

= 4 u^{2} - 4 - 2 u

4 u^{2} - 4 - 2 u = u^{2} + 1

Move

4

to the right side

4 u^{2} - 4 - 2 u = u^{2} + 1

Add

4

to both sides

4 u^{2} - 4 - 2 u + 4 = u^{2} + 1 + 4

Simplify

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

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

Solve

4 u^{2} - 2 u = u^{2} + 5 : u = \frac{5}{3}, u = - 1

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

Move

5

to the left side

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

Subtract

5

from both sides

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

Simplify

4 u^{2} - 2 u - 5 = u^{2}

4 u^{2} - 2 u - 5 = u^{2}

Move

u^{2}

to the left side

4 u^{2} - 2 u - 5 = u^{2}

Subtract

u^{2}

from both sides

4 u^{2} - 2 u - 5 - u^{2} = u^{2} - u^{2}

Simplify

3 u^{2} - 2 u - 5 = 0

3 u^{2} - 2 u - 5 = 0

Solve with the quadratic formula

3 u^{2} - 2 u - 5 = 0

Quadratic Equation Formula:

For

a = 3, b = - 2, c = - 5

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 5) = 8

(- 2)^{2} - 4 \cdot 3 (- 5)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 3 \cdot 5

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 3 \cdot 5

Multiply the numbers:

4 \cdot 3 \cdot 5 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 3}, u_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 3}

u = \frac{- ( - 2 ) + 8}{2 \cdot 3} : \frac{5}{3}

\frac{- ( - 2 ) + 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 + 8}{2 \cdot 3}

Add the numbers:

2 + 8 = 10

= \frac{10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{10}{6}

Cancel the common factor:

2

= \frac{5}{3}

u = \frac{- ( - 2 ) - 8}{2 \cdot 3} : - 1

\frac{- ( - 2 ) - 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{2 - 8}{2 \cdot 3}

Subtract the numbers:

2 - 8 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = - 1

u = \frac{5}{3}, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{5}{3}, u = - 1

u = \frac{5}{3}, u = - 1

Substitute back

u = e^{x},

solve for

x

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

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

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

Solve

e^{x} = - 1 :

No Solution for

x \in R

e^{x} = - 1

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

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

Verify Solutions:

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

True

Check the solutions by plugging them into

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

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

Plug in

x = ln (\frac{5}{3}) :

True

4 \cdot \frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} - \frac{1}{\frac{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}{2}} = 1

4 \cdot \frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} - \frac{1}{\frac{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}{2}} = 1

4 \cdot \frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} - \frac{1}{\frac{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}{2}}

4 \cdot \frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} = \frac{32}{17}

4 \cdot \frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}

\frac{e ^{l n (\frac{5}{3})} - e ^{- l n (\frac{5}{3})}}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} = \frac{8}{17}

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

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

e^{l n (\frac{5}{3})}

Apply log rule:

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

= \frac{5}{3}

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

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

Apply exponent rule:

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

= (e^{l n (\frac{5}{3})})^{- 1}

Apply log rule:

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

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

= (\frac{5}{3})^{- 1}

Apply exponent rule:

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

= \frac{1}{\frac{5}{3}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{3}{5}

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

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

e^{l n (\frac{5}{3})}

Apply log rule:

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

= \frac{5}{3}

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

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

Apply exponent rule:

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

= (e^{l n (\frac{5}{3})})^{- 1}

Apply log rule:

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

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

= (\frac{5}{3})^{- 1}

Apply exponent rule:

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

= \frac{1}{\frac{5}{3}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{3}{5}

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

Join

\frac{5}{3} + \frac{3}{5} : \frac{34}{15}

\frac{5}{3} + \frac{3}{5}

Least Common Multiplier of

3, 5 : 15

3, 5

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Multiply each factor the greatest number of times it occurs in either

3

5

= 3 \cdot 5

Multiply the numbers:

3 \cdot 5 = 15

= 15

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

15

For

\frac{5}{3} :

multiply the denominator and numerator by

5

\frac{5}{3} = \frac{5 \cdot 5}{3 \cdot 5} = \frac{25}{15}

For

\frac{3}{5} :

multiply the denominator and numerator by

3

\frac{3}{5} = \frac{3 \cdot 3}{5 \cdot 3} = \frac{9}{15}

= \frac{25}{15} + \frac{9}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{25 + 9}{15}

Add the numbers:

25 + 9 = 34

= \frac{34}{15}

= \frac{\frac{5}{3} - \frac{3}{5}}{\frac{34}{15}}

Join

\frac{5}{3} - \frac{3}{5} : \frac{16}{15}

\frac{5}{3} - \frac{3}{5}

Least Common Multiplier of

3, 5 : 15

3, 5

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Multiply each factor the greatest number of times it occurs in either

3

5

= 3 \cdot 5

Multiply the numbers:

3 \cdot 5 = 15

= 15

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

15

For

\frac{5}{3} :

multiply the denominator and numerator by

5

\frac{5}{3} = \frac{5 \cdot 5}{3 \cdot 5} = \frac{25}{15}

For

\frac{3}{5} :

multiply the denominator and numerator by

3

\frac{3}{5} = \frac{3 \cdot 3}{5 \cdot 3} = \frac{9}{15}

= \frac{25}{15} - \frac{9}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{25 - 9}{15}

Subtract the numbers:

25 - 9 = 16

= \frac{16}{15}

= \frac{\frac{16}{15}}{\frac{34}{15}}

Divide fractions:

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

= \frac{16 \cdot 15}{15 \cdot 34}

Cancel the common factor:

15

= \frac{16}{34}

Cancel the common factor:

2

= \frac{8}{17}

= 4 \cdot \frac{8}{17}

Multiply fractions:

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

= \frac{8 \cdot 4}{17}

Multiply the numbers:

8 \cdot 4 = 32

= \frac{32}{17}

\frac{1}{\frac{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}{2}} = \frac{15}{17}

\frac{1}{\frac{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}}{2}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

e^{l n (\frac{5}{3})}

Apply log rule:

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

= \frac{5}{3}

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

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

Apply exponent rule:

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

= (e^{l n (\frac{5}{3})})^{- 1}

Apply log rule:

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

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

= (\frac{5}{3})^{- 1}

Apply exponent rule:

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

= \frac{1}{\frac{5}{3}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{3}{5}

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

Join

\frac{5}{3} + \frac{3}{5} : \frac{34}{15}

\frac{5}{3} + \frac{3}{5}

Least Common Multiplier of

3, 5 : 15

3, 5

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Multiply each factor the greatest number of times it occurs in either

3

5

= 3 \cdot 5

Multiply the numbers:

3 \cdot 5 = 15

= 15

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

15

For

\frac{5}{3} :

multiply the denominator and numerator by

5

\frac{5}{3} = \frac{5 \cdot 5}{3 \cdot 5} = \frac{25}{15}

For

\frac{3}{5} :

multiply the denominator and numerator by

3

\frac{3}{5} = \frac{3 \cdot 3}{5 \cdot 3} = \frac{9}{15}

= \frac{25}{15} + \frac{9}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{25 + 9}{15}

Add the numbers:

25 + 9 = 34

= \frac{34}{15}

= \frac{2}{\frac{34}{15}}

Apply the fraction rule:

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

= \frac{2 \cdot 15}{34}

Multiply the numbers:

2 \cdot 15 = 30

= \frac{30}{34}

Cancel the common factor:

2

= \frac{15}{17}

= \frac{32}{17} - \frac{15}{17}

Simplify

\frac{32}{17} - \frac{15}{17}

Apply rule

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

= \frac{32 - 15}{17}

Subtract the numbers:

32 - 15 = 17

= \frac{17}{17}

Apply rule

\frac{a}{a} = 1

= 1

= 1

1 = 1

True

The solution is

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

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4tanh(x)-1/(cosh(x))=1 ?
The general solution for 4tanh(x)-1/(cosh(x))=1 is x=ln(5/3)