What is the general solution for tanh(x)+4sech(x)=4 ?

The general solution for tanh(x)+4sech(x)=4 is x=0,x=ln(5/3)

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tanh(x)+4sech(x)=4

Solution

tanh (x) + 4 sech (x) = 4

Solution

x = 0, x = ln (\frac{5}{3})

Degrees

x = 0^{\circ}, x = 29.26815 \dots^{\circ}

Solution steps

tanh (x) + 4 sech (x) = 4

Rewrite using trig identities

tanh (x) + 4 sech (x) = 4

Use the Hyperbolic identity:

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

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} + 4 sech (x) = 4

Use the Hyperbolic identity:

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

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

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

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} + 4 \cdot \frac{2}{e ^{x} + e ^{- x}} = 4 : x = 0, x = ln (\frac{5}{3})

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

Multiply both sides by

e^{x} + e^{- x}

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

Simplify

e^{x} - e^{- x} + 8 = 4 (e^{x} + e^{- x})

Apply exponent rules

e^{x} - e^{- x} + 8 = 4 (e^{x} + e^{- x})

Apply exponent rule:

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

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

e^{x} - (e^{x})^{- 1} + 8 = 4 (e^{x} + (e^{x})^{- 1})

e^{x} - (e^{x})^{- 1} + 8 = 4 (e^{x} + (e^{x})^{- 1})

Rewrite the equation with

e^{x} = u

u - (u)^{- 1} + 8 = 4 (u + (u)^{- 1})

Solve

u - u^{- 1} + 8 = 4 (u + u^{- 1}) : u = 1, u = \frac{5}{3}

u - u^{- 1} + 8 = 4 (u + u^{- 1})

Refine

u - \frac{1}{u} + 8 = 4 (u + \frac{1}{u})

Multiply both sides by

u

u - \frac{1}{u} + 8 = 4 (u + \frac{1}{u})

Multiply both sides by

u

uu - \frac{1}{u} u + 8 u = 4 (u + \frac{1}{u}) u

Simplify

uu - \frac{1}{u} u + 8 u = 4 (u + \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

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

Expand

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

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

= 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

u^{2} - 1 + 8 u = 4 u^{2} + 4

Move

1

to the right side

u^{2} - 1 + 8 u = 4 u^{2} + 4

Add

1

to both sides

u^{2} - 1 + 8 u + 1 = 4 u^{2} + 4 + 1

Simplify

u^{2} + 8 u = 4 u^{2} + 5

u^{2} + 8 u = 4 u^{2} + 5

Solve

u^{2} + 8 u = 4 u^{2} + 5 : u = 1, u = \frac{5}{3}

u^{2} + 8 u = 4 u^{2} + 5

Move

5

to the left side

u^{2} + 8 u = 4 u^{2} + 5

Subtract

5

from both sides

u^{2} + 8 u - 5 = 4 u^{2} + 5 - 5

Simplify

u^{2} + 8 u - 5 = 4 u^{2}

u^{2} + 8 u - 5 = 4 u^{2}

Move

4 u^{2}

to the left side

u^{2} + 8 u - 5 = 4 u^{2}

Subtract

4 u^{2}

from both sides

u^{2} + 8 u - 5 - 4 u^{2} = 4 u^{2} - 4 u^{2}

Simplify

- 3 u^{2} + 8 u - 5 = 0

- 3 u^{2} + 8 u - 5 = 0

Solve with the quadratic formula

- 3 u^{2} + 8 u - 5 = 0

Quadratic Equation Formula:

For

a = - 3, b = 8, c = - 5

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

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

8^{2} - 4 (- 3) (- 5)

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 5 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

Subtract the numbers:

64 - 60 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 8 + 2 = - 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

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 8 - 2 = - 10

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 10}{- 6}

Apply the fraction rule:

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

= \frac{10}{6}

Cancel the common factor:

2

= \frac{5}{3}

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u - u^{- 1} + 8

and compare to zero

u = 0

Take the denominator(s) of

4 (u + u^{- 1})

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

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

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

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})

x = 0, x = ln (\frac{5}{3})

Verify Solutions:

x = 0

True

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

True

Check the solutions by plugging them into

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

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

Plug in

x = 0 :

True

\frac{e ^{0} - e ^{- 0}}{e ^{0} + e ^{- 0}} + 4 \cdot \frac{2}{e ^{0} + e ^{- 0}} = 4

\frac{e ^{0} - e ^{- 0}}{e ^{0} + e ^{- 0}} + 4 \cdot \frac{2}{e ^{0} + e ^{- 0}} = 4

\frac{e ^{0} - e ^{- 0}}{e ^{0} + e ^{- 0}} + 4 \cdot \frac{2}{e ^{0} + e ^{- 0}}

Apply rule

a^{0} = 1, a \neq = 0

e^{0} = 1, e^{- 0} = 1

= \frac{1 - 1}{1 + 1} + 4 \cdot \frac{2}{1 + 1}

\frac{1 - 1}{1 + 1} = 0

\frac{1 - 1}{1 + 1}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{1 + 1}

Add the numbers:

1 + 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

4 \cdot \frac{2}{1 + 1} = 4

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

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

\frac{2}{1 + 1}

Add the numbers:

1 + 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

= 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 4

= 0 + 4

Add the numbers:

0 + 4 = 4

= 4

4 = 4

True

Plug in

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

True

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

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

\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})}} + 4 \cdot \frac{2}{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{2}{e ^{l n (\frac{5}{3})} + e ^{- l n (\frac{5}{3})}} = \frac{60}{17}

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

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

\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}

= 4 \cdot \frac{15}{17}

Multiply fractions:

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

= \frac{15 \cdot 4}{17}

Multiply the numbers:

15 \cdot 4 = 60

= \frac{60}{17}

= \frac{8}{17} + \frac{60}{17}

Simplify

\frac{8}{17} + \frac{60}{17}

Apply rule

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

= \frac{8 + 60}{17}

Add the numbers:

8 + 60 = 68

= \frac{68}{17}

Divide the numbers:

\frac{68}{17} = 4

= 4

= 4

4 = 4

True

The solutions are

x = 0, x = ln (\frac{5}{3})

x = 0, x = ln (\frac{5}{3})

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Frequently Asked Questions (FAQ)

What is the general solution for tanh(x)+4sech(x)=4 ?
The general solution for tanh(x)+4sech(x)=4 is x=0,x=ln(5/3)