What is the general solution for 4cosh(x)=7 ?

The general solution for 4cosh(x)=7 is x=ln((7+sqrt(33))/4),x=ln((7-sqrt(33))/4)

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4cosh(x)=7

Solution

4 cosh (x) = 7

Solution

x = ln (\frac{7 + 33}{4}), x = ln (\frac{7 - 33}{4})

Degrees

x = 66.39494 \dots^{\circ}, x = - 66.39494 \dots^{\circ}

Solution steps

4 cosh (x) = 7

Rewrite using trig identities

4 cosh (x) = 7

Use the Hyperbolic identity:

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

4 \cdot \frac{e ^{x} + e ^{- x}}{2} = 7

4 \cdot \frac{e ^{x} + e ^{- x}}{2} = 7

4 \cdot \frac{e ^{x} + e ^{- x}}{2} = 7 : x = ln (\frac{7 + 33}{4}), x = ln (\frac{7 - 33}{4})

4 \cdot \frac{e ^{x} + e ^{- x}}{2} = 7

Apply exponent rules

4 \cdot \frac{e ^{x} + e ^{- x}}{2} = 7

Apply exponent rule:

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

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

4 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} = 7

4 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} = 7

Rewrite the equation with

e^{x} = u

4 \cdot \frac{u + ( u ) ^{- 1}}{2} = 7

Solve

4 \cdot \frac{u + u ^{- 1}}{2} = 7 : u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

4 \cdot \frac{u + u ^{- 1}}{2} = 7

Refine

\frac{2 ( u ^{2} + 1 )}{u} = 7

Multiply both sides by

u

\frac{2 ( u ^{2} + 1 )}{u} = 7

Multiply both sides by

u

\frac{2 ( u ^{2} + 1 )}{u} u = 7 u

Simplify

2 (u^{2} + 1) = 7 u

2 (u^{2} + 1) = 7 u

Solve

2 (u^{2} + 1) = 7 u : u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

2 (u^{2} + 1) = 7 u

Expand

2 (u^{2} + 1) : 2 u^{2} + 2

2 (u^{2} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = u^{2}, c = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2} + 2

2 u^{2} + 2 = 7 u

Move

7 u

to the left side

2 u^{2} + 2 = 7 u

Subtract

7 u

from both sides

2 u^{2} + 2 - 7 u = 7 u - 7 u

Simplify

2 u^{2} + 2 - 7 u = 0

2 u^{2} + 2 - 7 u = 0

Write in the standard form

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

2 u^{2} - 7 u + 2 = 0

Solve with the quadratic formula

2 u^{2} - 7 u + 2 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 2

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 2 = 33

(- 7)^{2} - 4 \cdot 2 \cdot 2

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 7^{2} - 16

7^{2} = 49

= 49 - 16

Subtract the numbers:

49 - 16 = 33

= 33

u_{1, 2} = \frac{- ( - 7 ) \pm 33}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 33}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 33}{2 \cdot 2}

u = \frac{- ( - 7 ) + 33}{2 \cdot 2} : \frac{7 + 33}{4}

\frac{- ( - 7 ) + 33}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 33}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{7 + 33}{4}

u = \frac{- ( - 7 ) - 33}{2 \cdot 2} : \frac{7 - 33}{4}

\frac{- ( - 7 ) - 33}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 33}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{7 - 33}{4}

The solutions to the quadratic equation are:

u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

4 \frac{u + u ^{- 1}}{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

u = \frac{7 + 33}{4}, u = \frac{7 - 33}{4}

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{7 + 33}{4} : x = ln (\frac{7 + 33}{4})

e^{x} = \frac{7 + 33}{4}

Apply exponent rules

e^{x} = \frac{7 + 33}{4}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{7 + 33}{4})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{7 + 33}{4})

x = ln (\frac{7 + 33}{4})

Solve

e^{x} = \frac{7 - 33}{4} : x = ln (\frac{7 - 33}{4})

e^{x} = \frac{7 - 33}{4}

Apply exponent rules

e^{x} = \frac{7 - 33}{4}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{7 - 33}{4})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{7 - 33}{4})

x = ln (\frac{7 - 33}{4})

x = ln (\frac{7 + 33}{4}), x = ln (\frac{7 - 33}{4})

x = ln (\frac{7 + 33}{4}), x = ln (\frac{7 - 33}{4})

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Frequently Asked Questions (FAQ)

What is the general solution for 4cosh(x)=7 ?
The general solution for 4cosh(x)=7 is x=ln((7+sqrt(33))/4),x=ln((7-sqrt(33))/4)