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

The general solution for solvefor u,x=5tanh(u) is u= 1/2 ln((x+5)/(5-x)),u=ln(-sqrt((x+5)/(5-x)))

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solvefor u,x=5tanh(u)

Solution

solve for

u, x = 5 tanh (u)

Solution

u = \frac{1}{2} ln (\frac{x + 5}{5 - x}), u = ln (- \frac{x + 5}{5 - x})

Solution steps

x = 5 tanh (u)

Switch sides

5 tanh (u) = x

Rewrite using trig identities

5 tanh (u) = x

Use the Hyperbolic identity:

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

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

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

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

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{u} = v

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

Solve

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

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

Refine

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

Multiply both sides by

v^{2} + 1

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

Multiply both sides by

v^{2} + 1

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

Simplify

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

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

Solve

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

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

Move

x (v^{2} + 1)

to the left side

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

Subtract

x (v^{2} + 1)

from both sides

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

Simplify

5 (v^{2} - 1) - x (v^{2} + 1) = 0

5 (v^{2} - 1) - x (v^{2} + 1) = 0

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 - x (v^{2} + 1) = 0

Expand

- x (v^{2} + 1) : - x v^{2} - x

- x (v^{2} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = - x, b = v^{2}, c = 1

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

Apply minus-plus rules

+ (- a) = - a

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

Multiply:

1 \cdot x = x

= - x v^{2} - x

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

Move

x

to the right side

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

Add

x

to both sides

5 v^{2} - 5 - x v^{2} - x + x = 0 + x

Simplify

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

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

Move

5

to the right side

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

Add

5

to both sides

5 v^{2} - 5 - x v^{2} + 5 = x + 5

Simplify

5 v^{2} - x v^{2} = x + 5

5 v^{2} - x v^{2} = x + 5

Factor

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

5 v^{2} - x v^{2}

Factor out common term

v^{2}

= v^{2} (5 - x)

v^{2} (5 - x) = x + 5

Divide both sides by

5 - x

v^{2} (5 - x) = x + 5

Divide both sides by

5 - x

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

Simplify

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

Simplify

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

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

Cancel the common factor:

5 - x

= v^{2}

Simplify

\frac{x}{5 - x} + \frac{5}{5 - x} : \frac{x + 5}{5 - x}

\frac{x}{5 - x} + \frac{5}{5 - x}

Apply rule

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

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

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

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

v = \frac{x + 5}{5 - x}, v = - \frac{x + 5}{5 - x}

v = \frac{x + 5}{5 - x}, v = - \frac{x + 5}{5 - x}

v = \frac{x + 5}{5 - x}, v = - \frac{x + 5}{5 - x}

Substitute back

v = e^{u},

solve for

u

Solve

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

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

Apply exponent rules

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

Apply exponent rule:

a = a^{\frac{1}{2}}

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

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln ((\frac{x + 5}{5 - x})^{\frac{1}{2}})

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

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

u = \frac{1}{2} ln (\frac{x + 5}{5 - x})

u = \frac{1}{2} ln (\frac{x + 5}{5 - x})

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{u}) = u

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

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

u = \frac{1}{2} ln (\frac{x + 5}{5 - x}), u = ln (- \frac{x + 5}{5 - x})

u = \frac{1}{2} ln (\frac{x + 5}{5 - x}), u = ln (- \frac{x + 5}{5 - x})

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor u,x=5tanh(u) ?
The general solution for solvefor u,x=5tanh(u) is u= 1/2 ln((x+5)/(5-x)),u=ln(-sqrt((x+5)/(5-x)))