What is the general solution for solvefor u,x=5sinh(u) ?

The general solution for solvefor u,x=5sinh(u) is u=ln((x+sqrt(x^2+25))/5),u=ln((x-sqrt(x^2+25))/5)

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Popular Trigonometry >

solvefor u,x=5sinh(u)

Solution

solve for

u, x = 5 sinh (u)

Solution

u = ln (\frac{x + x ^{2} + 25}{5}), u = ln (\frac{x - x ^{2} + 25}{5})

Solution steps

x = 5 sinh (u)

Switch sides

5 sinh (u) = x

Rewrite using trig identities

5 sinh (u) = x

Use the Hyperbolic identity:

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

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

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

5 \cdot \frac{e ^{u} - e ^{- u}}{2} = x : u = ln (\frac{x + x ^{2} + 25}{5}), u = ln (\frac{x - x ^{2} + 25}{5})

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{u} = v

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

Solve

5 \cdot \frac{v - v ^{- 1}}{2} = x : v = \frac{x + x ^{2} + 25}{5}, v = \frac{x - x ^{2} + 25}{5}

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

Refine

\frac{5 ( v ^{2} - 1 )}{2 v} = x

Multiply both sides by

v

\frac{5 ( v ^{2} - 1 )}{2 v} = x

Multiply both sides by

v

\frac{5 ( v ^{2} - 1 )}{2 v} v = xv

Simplify

\frac{5 ( v ^{2} - 1 )}{2} = xv

\frac{5 ( v ^{2} - 1 )}{2} = xv

Solve

\frac{5 ( v ^{2} - 1 )}{2} = xv : v = \frac{x + x ^{2} + 25}{5}, v = \frac{x - x ^{2} + 25}{5}

\frac{5 ( v ^{2} - 1 )}{2} = xv

Multiply both sides by

2

\frac{5 ( v ^{2} - 1 )}{2} = xv

Multiply both sides by

2

\frac{5 ( v ^{2} - 1 )}{2} \cdot 2 = xv \cdot 2

Simplify

5 (v^{2} - 1) = 2 xv

5 (v^{2} - 1) = 2 xv

Expand

5 (v^{2} - 1) : 5 v^{2} - 5

5 (v^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = v^{2}, c = 1

= 5 v^{2} - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 v^{2} - 5

5 v^{2} - 5 = 2 xv

Move

2 xv

to the left side

5 v^{2} - 5 = 2 xv

Subtract

2 xv

from both sides

5 v^{2} - 5 - 2 xv = 2 xv - 2 xv

Simplify

5 v^{2} - 5 - 2 xv = 0

5 v^{2} - 5 - 2 xv = 0

Write in the standard form

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

5 v^{2} - 2 xv - 5 = 0

Solve with the quadratic formula

5 v^{2} - 2 xv - 5 = 0

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 2 x)^{2} - 4 \cdot 5 (- 5) : 2 x^{2} + 25

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

Apply rule

- (- a) = a

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

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

(- 2 x)^{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 5 \cdot 5 = 100

4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 100

= 2^{2} x^{2} + 100

Factor

2^{2} x^{2} + 100 : 4 (x^{2} + 25)

2^{2} x^{2} + 100

Rewrite as

= 4 x^{2} + 4 \cdot 25

Factor out common term

4

= 4 (x^{2} + 25)

= 4 (x^{2} + 25)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 x^{2} + 25

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 x^{2} + 25

v_{1, 2} = \frac{- ( - 2 x ) \pm 2 x ^{2} + 25}{2 \cdot 5}

Separate the solutions

v_{1} = \frac{- ( - 2 x ) + 2 x ^{2} + 25}{2 \cdot 5}, v_{2} = \frac{- ( - 2 x ) - 2 x ^{2} + 25}{2 \cdot 5}

v = \frac{- ( - 2 x ) + 2 x ^{2} + 25}{2 \cdot 5} : \frac{x + x ^{2} + 25}{5}

\frac{- ( - 2 x ) + 2 x ^{2} + 25}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{2 x + 2 x ^{2} + 25}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2 x + 2 x ^{2} + 25}{10}

Factor out common term

2

= \frac{2 ( x + x ^{2} + 25 )}{10}

Cancel the common factor:

2

= \frac{x + x ^{2} + 25}{5}

v = \frac{- ( - 2 x ) - 2 x ^{2} + 25}{2 \cdot 5} : \frac{x - x ^{2} + 25}{5}

\frac{- ( - 2 x ) - 2 x ^{2} + 25}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{2 x - 2 x ^{2} + 25}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{2 x - 2 x ^{2} + 25}{10}

Factor out common term

2

= \frac{2 ( x - x ^{2} + 25 )}{10}

Cancel the common factor:

2

= \frac{x - x ^{2} + 25}{5}

The solutions to the quadratic equation are:

v = \frac{x + x ^{2} + 25}{5}, v = \frac{x - x ^{2} + 25}{5}

v = \frac{x + x ^{2} + 25}{5}, v = \frac{x - x ^{2} + 25}{5}

v = \frac{x + x ^{2} + 25}{5}, v = \frac{x - x ^{2} + 25}{5}

Substitute back

v = e^{u},

solve for

u

Solve

e^{u} = \frac{x + x ^{2} + 25}{5} : u = ln (\frac{x + x ^{2} + 25}{5})

e^{u} = \frac{x + x ^{2} + 25}{5}

Apply exponent rules

e^{u} = \frac{x + x ^{2} + 25}{5}

f (x) = g (x)

, then

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

ln (e^{u}) = ln (\frac{x + x ^{2} + 25}{5})

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (\frac{x + x ^{2} + 25}{5})

u = ln (\frac{x + x ^{2} + 25}{5})

Solve

e^{u} = \frac{x - x ^{2} + 25}{5} : u = ln (\frac{x - x ^{2} + 25}{5})

e^{u} = \frac{x - x ^{2} + 25}{5}

Apply exponent rules

e^{u} = \frac{x - x ^{2} + 25}{5}

f (x) = g (x)

, then

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

ln (e^{u}) = ln (\frac{x - x ^{2} + 25}{5})

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (\frac{x - x ^{2} + 25}{5})

u = ln (\frac{x - x ^{2} + 25}{5})

u = ln (\frac{x + x ^{2} + 25}{5}), u = ln (\frac{x - x ^{2} + 25}{5})

u = ln (\frac{x + x ^{2} + 25}{5}), u = ln (\frac{x - x ^{2} + 25}{5})

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor u,x=5sinh(u) ?
The general solution for solvefor u,x=5sinh(u) is u=ln((x+sqrt(x^2+25))/5),u=ln((x-sqrt(x^2+25))/5)