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

The general solution for cosh(x)=4 is x=ln(4+sqrt(15)),x=ln(4-sqrt(15))

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cosh(x)=4

Solution

cosh (x) = 4

Solution

x = ln (4 + 15), x = ln (4 - 15)

Degrees

x = 118.22623 \dots^{\circ}, x = - 118.22623 \dots^{\circ}

Solution steps

cosh (x) = 4

Rewrite using trig identities

cosh (x) = 4

Use the Hyperbolic identity:

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

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

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

\frac{e ^{x} + e ^{- x}}{2} = 4 : x = ln (4 + 15), x = ln (4 - 15)

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

Multiply both sides by

2

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

Simplify

e^{x} + e^{- x} = 8

Apply exponent rules

e^{x} + e^{- x} = 8

Apply exponent rule:

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

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

e^{x} + (e^{x})^{- 1} = 8

e^{x} + (e^{x})^{- 1} = 8

Rewrite the equation with

e^{x} = u

u + (u)^{- 1} = 8

Solve

u + u^{- 1} = 8 : u = 4 + 15, u = 4 - 15

u + u^{- 1} = 8

Refine

u + \frac{1}{u} = 8

Multiply both sides by

u

u + \frac{1}{u} = 8

Multiply both sides by

u

uu + \frac{1}{u} u = 8 u

Simplify

uu + \frac{1}{u} u = 8 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 = 8 u

u^{2} + 1 = 8 u

u^{2} + 1 = 8 u

Solve

u^{2} + 1 = 8 u : u = 4 + 15, u = 4 - 15

u^{2} + 1 = 8 u

Move

8 u

to the left side

u^{2} + 1 = 8 u

Subtract

8 u

from both sides

u^{2} + 1 - 8 u = 8 u - 8 u

Simplify

u^{2} + 1 - 8 u = 0

u^{2} + 1 - 8 u = 0

Write in the standard form

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

u^{2} - 8 u + 1 = 0

Solve with the quadratic formula

u^{2} - 8 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 8, c = 1

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 1 = 215

(- 8)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 8^{2} - 4

8^{2} = 64

= 64 - 4

Subtract the numbers:

64 - 4 = 60

= 60

Prime factorization of

60 : 2^{2} \cdot 3 \cdot 5

60

60

divides by

260 = 30 \cdot 2

= 2 \cdot 30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

Apply radical rule:

= 2^{2} 3 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 3 \cdot 5

Refine

= 215

u_{1, 2} = \frac{- ( - 8 ) \pm 2 15}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 2 15}{2 \cdot 1}, u_{2} = \frac{- ( - 8 ) - 2 15}{2 \cdot 1}

u = \frac{- ( - 8 ) + 2 15}{2 \cdot 1} : 4 + 15

\frac{- ( - 8 ) + 2 15}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 + 2 15}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8 + 2 15}{2}

Factor

8 + 215 : 2 (4 + 15)

8 + 215

Rewrite as

= 2 \cdot 4 + 215

Factor out common term

2

= 2 (4 + 15)

= \frac{2 ( 4 + 15 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 4 + 15

u = \frac{- ( - 8 ) - 2 15}{2 \cdot 1} : 4 - 15

\frac{- ( - 8 ) - 2 15}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{8 - 2 15}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8 - 2 15}{2}

Factor

8 - 215 : 2 (4 - 15)

8 - 215

Rewrite as

= 2 \cdot 4 - 215

Factor out common term

2

= 2 (4 - 15)

= \frac{2 ( 4 - 15 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 4 - 15

The solutions to the quadratic equation are:

u = 4 + 15, u = 4 - 15

u = 4 + 15, u = 4 - 15

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 = 4 + 15, u = 4 - 15

u = 4 + 15, u = 4 - 15

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 4 + 15 : x = ln (4 + 15)

e^{x} = 4 + 15

Apply exponent rules

e^{x} = 4 + 15

f (x) = g (x)

, then

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

ln (e^{x}) = ln (4 + 15)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (4 + 15)

x = ln (4 + 15)

Solve

e^{x} = 4 - 15 : x = ln (4 - 15)

e^{x} = 4 - 15

Apply exponent rules

e^{x} = 4 - 15

f (x) = g (x)

, then

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

ln (e^{x}) = ln (4 - 15)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (4 - 15)

x = ln (4 - 15)

x = ln (4 + 15), x = ln (4 - 15)

x = ln (4 + 15), x = ln (4 - 15)

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Frequently Asked Questions (FAQ)

What is the general solution for cosh(x)=4 ?
The general solution for cosh(x)=4 is x=ln(4+sqrt(15)),x=ln(4-sqrt(15))