What is the general solution for sinh(x)=1 ?

The general solution for sinh(x)=1 is x=ln(1+sqrt(2))

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sinh(x)=1

Solution

sinh (x) = 1

Solution

x = ln (1 + 2)

Degrees

x = 50.49898 \dots^{\circ}

Solution steps

sinh (x) = 1

Rewrite using trig identities

sinh (x) = 1

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

2

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

Simplify

e^{x} - e^{- x} = 2

Apply exponent rules

e^{x} - e^{- x} = 2

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

u - u^{- 1} = 2

Refine

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

u^{2} - 1 = 2 u

u^{2} - 1 = 2 u

Solve

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

u^{2} - 1 = 2 u

Move

2 u

to the left side

u^{2} - 1 = 2 u

Subtract

2 u

from both sides

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

Simplify

u^{2} - 1 - 2 u = 0

u^{2} - 1 - 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - 2 u - 1 = 0

Solve with the quadratic formula

u^{2} - 2 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

(- 2)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

u = 1 + 2, u = 1 - 2

u = 1 + 2, u = 1 - 2

Verify Solutions

Find undefined (singularity) points:

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 = 1 + 2, u = 1 - 2

u = 1 + 2, u = 1 - 2

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 1 + 2 : x = ln (1 + 2)

e^{x} = 1 + 2

Apply exponent rules

e^{x} = 1 + 2

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1 + 2)

x = ln (1 + 2)

Solve

e^{x} = 1 - 2 :

No Solution for

x \in R

e^{x} = 1 - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (1 + 2)

x = ln (1 + 2)

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Frequently Asked Questions (FAQ)

What is the general solution for sinh(x)=1 ?
The general solution for sinh(x)=1 is x=ln(1+sqrt(2))