What is the general solution for cosh(x-1)=2 ?

The general solution for cosh(x-1)=2 is x=ln(e(2+sqrt(3))),x=ln(e(2-sqrt(3)))

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cosh(x-1)=2

Solution

cosh (x - 1) = 2

Solution

x = ln (e (2 + 3)), x = ln (e (2 - 3))

Degrees

x = 132.75190 \dots^{\circ}, x = - 18.16034 \dots^{\circ}

Solution steps

cosh (x - 1) = 2

Rewrite using trig identities

cosh (x - 1) = 2

Use the Hyperbolic identity:

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

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

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

\frac{e ^{x - 1} + e ^{- (x - 1)}}{2} = 2 : x = ln (e (2 + 3)), x = ln (e (2 - 3))

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

Multiply both sides by

2

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

Simplify

e^{x - 1} + e^{- (x - 1)} = 4

Apply exponent rules

e^{x - 1} + e^{- (x - 1)} = 4

Apply exponent rule:

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

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

e^{x} e^{- 1} + e^{- 1 \cdot x} e^{1} = 4

Apply exponent rule:

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

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

e^{x} e^{- 1} + (e^{x})^{- 1} e^{1} = 4

e^{x} e^{- 1} + (e^{x})^{- 1} e^{1} = 4

Rewrite the equation with

e^{x} = u

u e^{- 1} + (u)^{- 1} e^{1} = 4

Solve

u e^{- 1} + u^{- 1} e^{1} = 4 : u = e (2 + 3), u = e (2 - 3)

u e^{- 1} + u^{- 1} e^{1} = 4

Refine

\frac{1}{e} u + \frac{e}{u} = 4

Multiply by LCM

\frac{1}{e} u + \frac{e}{u} = 4

Find Least Common Multiplier of

e, u : e u

e, u

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

e

u

= e u

Multiply by LCM=

e u

\frac{1}{e} u e u + \frac{e}{u} e u = 4 e u

Simplify

\frac{1}{e} u e u + \frac{e}{u} e u = 4 e u

Simplify

\frac{1}{e} u e u : u^{2}

\frac{1}{e} u e u

Apply exponent rule:

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

uu = u^{1 + 1}

= \frac{1}{e} e u^{1 + 1}

Add the numbers:

1 + 1 = 2

= \frac{1}{e} e u^{2}

Multiply fractions:

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

= \frac{1 e}{e} u^{2}

Cancel the common factor:

e

= u^{2} \cdot 1

Multiply:

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

= u^{2}

Simplify

\frac{e}{u} e u : e^{2}

\frac{e}{u} e u

Multiply fractions:

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

= \frac{ee u}{u}

Cancel the common factor:

u

= ee

Apply exponent rule:

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

ee = e^{1 + 1}

= e^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{2}

u^{2} + e^{2} = 4 e u

u^{2} + e^{2} = 4 e u

u^{2} + e^{2} = 4 e u

Solve

u^{2} + e^{2} = 4 e u : u = e (2 + 3), u = e (2 - 3)

u^{2} + e^{2} = 4 e u

Move

4 e u

to the left side

u^{2} + e^{2} = 4 e u

Subtract

4 e u

from both sides

u^{2} + e^{2} - 4 e u = 4 e u - 4 e u

Simplify

u^{2} + e^{2} - 4 e u = 0

u^{2} + e^{2} - 4 e u = 0

Write in the standard form

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

u^{2} - 4 e u + e^{2} = 0

Solve with the quadratic formula

u^{2} - 4 e u + e^{2} = 0

Quadratic Equation Formula:

For

a = 1, b = - 4 e, c = e^{2}

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

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

(- 4 e)^{2} - 4 \cdot 1 \cdot e^{2} = 23 e

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

(- 4 e)^{2} = 4^{2} e^{2}

(- 4 e)^{2}

Apply exponent rule:

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

n

is even

(- 4 e)^{2} = (4 e)^{2}

= (4 e)^{2}

Apply exponent rule:

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

= 4^{2} e^{2}

4 \cdot 1 \cdot e^{2} = 4 e^{2}

4 \cdot 1 \cdot e^{2}

Multiply the numbers:

4 \cdot 1 = 4

= 4 e^{2}

= 4^{2} e^{2} - 4 e^{2}

4^{2} = 16

= 16 e^{2} - 4 e^{2}

Add similar elements:

16 e^{2} - 4 e^{2} = 12 e^{2}

= 12 e^{2}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 12 e^{2}

12 = 23

12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= 23 e^{2}

Apply radical rule:

assuming

a \geq 0

e^{2} = e

= 23 e

u_{1, 2} = \frac{- ( - 4 e ) \pm 2 3 e}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 e ) + 2 3 e}{2 \cdot 1}, u_{2} = \frac{- ( - 4 e ) - 2 3 e}{2 \cdot 1}

u = \frac{- ( - 4 e ) + 2 3 e}{2 \cdot 1} : e (2 + 3)

\frac{- ( - 4 e ) + 2 3 e}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 e + 2 3 e}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 e + 2 3 e}{2}

Factor

4 e + 23 e : 2 e (2 + 3)

4 e + 23 e

Rewrite as

= 2 \cdot 2 e + 2 e 3

Factor out common term

2 e

= 2 e (2 + 3)

= \frac{2 e ( 2 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= e (2 + 3)

u = \frac{- ( - 4 e ) - 2 3 e}{2 \cdot 1} : e (2 - 3)

\frac{- ( - 4 e ) - 2 3 e}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 e - 2 3 e}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 e - 2 3 e}{2}

Factor

4 e - 23 e : 2 e (2 - 3)

4 e - 23 e

Rewrite as

= 2 \cdot 2 e - 2 e 3

Factor out common term

2 e

= 2 e (2 - 3)

= \frac{2 e ( 2 - 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= e (2 - 3)

The solutions to the quadratic equation are:

u = e (2 + 3), u = e (2 - 3)

u = e (2 + 3), u = e (2 - 3)

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u e^{- 1} + u^{- 1} e^{1}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = e (2 + 3), u = e (2 - 3)

u = e (2 + 3), u = e (2 - 3)

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = e (2 + 3) : x = ln (e (2 + 3))

e^{x} = e (2 + 3)

Apply exponent rules

e^{x} = e (2 + 3)

f (x) = g (x)

, then

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

ln (e^{x}) = ln (e (2 + 3))

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (e (2 + 3))

x = ln (e (2 + 3))

Solve

e^{x} = e (2 - 3) : x = ln (e (2 - 3))

e^{x} = e (2 - 3)

Apply exponent rules

e^{x} = e (2 - 3)

f (x) = g (x)

, then

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

ln (e^{x}) = ln (e (2 - 3))

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (e (2 - 3))

x = ln (e (2 - 3))

x = ln (e (2 + 3)), x = ln (e (2 - 3))

x = ln (e (2 + 3)), x = ln (e (2 - 3))

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Frequently Asked Questions (FAQ)

What is the general solution for cosh(x-1)=2 ?
The general solution for cosh(x-1)=2 is x=ln(e(2+sqrt(3))),x=ln(e(2-sqrt(3)))