What is the general solution for cosh(x)= 5/4 ?

The general solution for cosh(x)= 5/4 is x=ln(2),x=-ln(2)

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Popular Trigonometry >

cosh(x)= 5/4

Solution

cosh (x) = \frac{5}{4}

Solution

x = ln (2), x = - ln (2)

Degrees

x = 39.71440 \dots^{\circ}, x = - 39.71440 \dots^{\circ}

Solution steps

cosh (x) = \frac{5}{4}

Rewrite using trig identities

cosh (x) = \frac{5}{4}

Use the Hyperbolic identity:

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

\frac{e ^{x} + e ^{- x}}{2} = \frac{5}{4}

\frac{e ^{x} + e ^{- x}}{2} = \frac{5}{4}

\frac{e ^{x} + e ^{- x}}{2} = \frac{5}{4} : x = ln (2), x = - ln (2)

\frac{e ^{x} + e ^{- x}}{2} = \frac{5}{4}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

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

Simplify

(e^{x} + e^{- x}) \cdot 4 = 10

Apply exponent rules

(e^{x} + e^{- x}) \cdot 4 = 10

Apply exponent rule:

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

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

(e^{x} + (e^{x})^{- 1}) \cdot 4 = 10

(e^{x} + (e^{x})^{- 1}) \cdot 4 = 10

Rewrite the equation with

e^{x} = u

(u + (u)^{- 1}) \cdot 4 = 10

Solve

(u + u^{- 1}) \cdot 4 = 10 : u = 2, u = \frac{1}{2}

(u + u^{- 1}) \cdot 4 = 10

Refine

(u + \frac{1}{u}) \cdot 4 = 10

Simplify

(u + \frac{1}{u}) \cdot 4 : 4 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 4

Apply the commutative law:

(u + \frac{1}{u}) \cdot 4 = 4 (u + \frac{1}{u})

4 (u + \frac{1}{u})

4 (u + \frac{1}{u}) = 10

Expand

4 (u + \frac{1}{u}) : 4 u + \frac{4}{u}

4 (u + \frac{1}{u})

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = u, c = \frac{1}{u}

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

4 \cdot \frac{1}{u} = \frac{4}{u}

4 \cdot \frac{1}{u}

Multiply fractions:

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

= \frac{1 \cdot 4}{u}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{u}

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

4 u + \frac{4}{u} = 10

Multiply both sides by

u

4 u + \frac{4}{u} = 10

Multiply both sides by

u

4 uu + \frac{4}{u} u = 10 u

Simplify

4 uu + \frac{4}{u} u = 10 u

Simplify

4 uu : 4 u^{2}

4 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 u^{2}

Simplify

\frac{4}{u} u : 4

\frac{4}{u} u

Multiply fractions:

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

= \frac{4 u}{u}

Cancel the common factor:

u

= 4

4 u^{2} + 4 = 10 u

4 u^{2} + 4 = 10 u

4 u^{2} + 4 = 10 u

Solve

4 u^{2} + 4 = 10 u : u = 2, u = \frac{1}{2}

4 u^{2} + 4 = 10 u

Move

10 u

to the left side

4 u^{2} + 4 = 10 u

Subtract

10 u

from both sides

4 u^{2} + 4 - 10 u = 10 u - 10 u

Simplify

4 u^{2} + 4 - 10 u = 0

4 u^{2} + 4 - 10 u = 0

Write in the standard form

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

4 u^{2} - 10 u + 4 = 0

Solve with the quadratic formula

4 u^{2} - 10 u + 4 = 0

Quadratic Equation Formula:

For

a = 4, b = - 10, c = 4

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

(- 10)^{2} - 4 \cdot 4 \cdot 4 = 6

(- 10)^{2} - 4 \cdot 4 \cdot 4

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 1 0^{2} - 64

1 0^{2} = 100

= 100 - 64

Subtract the numbers:

100 - 64 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 10 ) \pm 6}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 6}{2 \cdot 4}, u_{2} = \frac{- ( - 10 ) - 6}{2 \cdot 4}

u = \frac{- ( - 10 ) + 6}{2 \cdot 4} : 2

\frac{- ( - 10 ) + 6}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{10 + 6}{2 \cdot 4}

Add the numbers:

10 + 6 = 16

= \frac{16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 10 ) - 6}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 10 ) - 6}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{10 - 6}{2 \cdot 4}

Subtract the numbers:

10 - 6 = 4

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u + u^{- 1}) 4

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = e^{x},

solve for

x

Solve

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

e^{x} = 2

Apply exponent rules

e^{x} = 2

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2)

x = ln (2)

Solve

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

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

Apply exponent rules

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

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{1}{2})

Simplify

ln (\frac{1}{2}) : - ln (2)

ln (\frac{1}{2})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (2)

x = - ln (2)

x = - ln (2)

x = ln (2), x = - ln (2)

x = ln (2), x = - ln (2)

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cosh(x)= 5/4 ?
The general solution for cosh(x)= 5/4 is x=ln(2),x=-ln(2)