What is the general solution for solvefor u,x=sinh(u) ?

The general solution for solvefor u,x=sinh(u) is u=ln(x+sqrt(x^2+1)),u=ln(x-sqrt(x^2+1))

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solvefor u,x=sinh(u)

Solution

solve for

u, x = sinh (u)

Solution

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

Solution steps

x = sinh (u)

Switch sides

sinh (u) = x

Rewrite using trig identities

sinh (u) = x

Use the Hyperbolic identity:

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

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

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

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

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

Multiply both sides by

2

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

Simplify

e^{u} - e^{- u} = x \cdot 2

Apply exponent rules

e^{u} - e^{- u} = x \cdot 2

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{u} = v

v - v^{- 1} = x \cdot 2

Solve

v - v^{- 1} = x \cdot 2 : v = x + x^{2} + 1, v = x - x^{2} + 1

v - v^{- 1} = x \cdot 2

Refine

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

Multiply both sides by

v

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

Multiply both sides by

v

vv - \frac{1}{v} v = x \cdot 2 v

Simplify

vv - \frac{1}{v} v = x \cdot 2 v

Simplify

vv : v^{2}

vv

Apply exponent rule:

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

vv = v^{1 + 1}

= v^{1 + 1}

Add the numbers:

1 + 1 = 2

= v^{2}

Simplify

- \frac{1}{v} v : - 1

- \frac{1}{v} v

Multiply fractions:

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

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

Cancel the common factor:

v

= - 1

v^{2} - 1 = x \cdot 2 v

v^{2} - 1 = x \cdot 2 v

v^{2} - 1 = x \cdot 2 v

Solve

v^{2} - 1 = x \cdot 2 v : v = x + x^{2} + 1, v = x - x^{2} + 1

v^{2} - 1 = x \cdot 2 v

Move

x 2 v

to the left side

v^{2} - 1 = x \cdot 2 v

Subtract

x 2 v

from both sides

v^{2} - 1 - x \cdot 2 v = x \cdot 2 v - x \cdot 2 v

Simplify

v^{2} - 1 - x \cdot 2 v = 0

v^{2} - 1 - x \cdot 2 v = 0

Write in the standard form

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

v^{2} - x \cdot 2 v - 1 = 0

Solve with the quadratic formula

v^{2} - x \cdot 2 v - 1 = 0

Quadratic Equation Formula:

For

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

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

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

Simplify

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

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

Apply rule

- (- a) = a

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

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

(- x \cdot 2)^{2}

Apply exponent rule:

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

n

is even

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

= (2 x)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} x^{2}

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 2^{2} x^{2} + 4

Factor

x^{2} 2^{2} + 4 : 4 (x^{2} + 1)

x^{2} \cdot 2^{2} + 4

Rewrite as

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

Factor out common term

4

= 4 (x^{2} + 1)

= 4 (x^{2} + 1)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 x^{2} + 1

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 x^{2} + 1

v_{1, 2} = \frac{- ( - x \cdot 2 ) \pm 2 x ^{2} + 1}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - x \cdot 2 ) + 2 x ^{2} + 1}{2 \cdot 1}, v_{2} = \frac{- ( - x \cdot 2 ) - 2 x ^{2} + 1}{2 \cdot 1}

v = \frac{- ( - x 2 ) + 2 x ^{2} + 1}{2 \cdot 1} : x + x^{2} + 1

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor out common term

2

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

Divide the numbers:

\frac{2}{2} = 1

= x + x^{2} + 1

v = \frac{- ( - x 2 ) - 2 x ^{2} + 1}{2 \cdot 1} : x - x^{2} + 1

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor out common term

2

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

Divide the numbers:

\frac{2}{2} = 1

= x - x^{2} + 1

The solutions to the quadratic equation are:

v = x + x^{2} + 1, v = x - x^{2} + 1

v = x + x^{2} + 1, v = x - x^{2} + 1

v = x + x^{2} + 1, v = x - x^{2} + 1

Substitute back

v = e^{u},

solve for

u

Solve

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

e^{u} = x + x^{2} + 1

Apply exponent rules

e^{u} = x + x^{2} + 1

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

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

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

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (x - x^{2} + 1)

u = ln (x - x^{2} + 1)

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor u,x=sinh(u) ?
The general solution for solvefor u,x=sinh(u) is u=ln(x+sqrt(x^2+1)),u=ln(x-sqrt(x^2+1))