What is the general solution for cosh(x)= 17/15 ?

The general solution for cosh(x)= 17/15 is x=ln(5/3),x=ln(3/5)

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Popular >

cosh(x)= 17/15

Solution

cosh (x) = \frac{17}{15}

Solution

x = ln (\frac{5}{3}), x = ln (\frac{3}{5})

Degrees

x = 29.26815 \dots^{\circ}, x = - 29.26815 \dots^{\circ}

Solution steps

cosh (x) = \frac{17}{15}

Rewrite using trig identities

cosh (x) = \frac{17}{15}

Use the Hyperbolic identity:

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

\frac{e ^{x} + e ^{- x}}{2} = \frac{17}{15}

\frac{e ^{x} + e ^{- x}}{2} = \frac{17}{15}

\frac{e ^{x} + e ^{- x}}{2} = \frac{17}{15} : x = ln (\frac{5}{3}), x = ln (\frac{3}{5})

\frac{e ^{x} + e ^{- x}}{2} = \frac{17}{15}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

(e^{x} + e^{- x}) \cdot 15 = 2 \cdot 17

Simplify

(e^{x} + e^{- x}) \cdot 15 = 34

Apply exponent rules

(e^{x} + e^{- x}) \cdot 15 = 34

Apply exponent rule:

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

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

(e^{x} + (e^{x})^{- 1}) \cdot 15 = 34

(e^{x} + (e^{x})^{- 1}) \cdot 15 = 34

Rewrite the equation with

e^{x} = u

(u + (u)^{- 1}) \cdot 15 = 34

Solve

(u + u^{- 1}) \cdot 15 = 34 : u = \frac{5}{3}, u = \frac{3}{5}

(u + u^{- 1}) \cdot 15 = 34

Refine

(u + \frac{1}{u}) \cdot 15 = 34

Simplify

(u + \frac{1}{u}) \cdot 15 : 15 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 15

Apply the commutative law:

(u + \frac{1}{u}) \cdot 15 = 15 (u + \frac{1}{u})

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

15 (u + \frac{1}{u}) = 34

Expand

15 (u + \frac{1}{u}) : 15 u + \frac{15}{u}

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

Apply the distributive law:

a (b + c) = ab + a c

a = 15, b = u, c = \frac{1}{u}

= 15 u + 15 \cdot \frac{1}{u}

15 \cdot \frac{1}{u} = \frac{15}{u}

15 \cdot \frac{1}{u}

Multiply fractions:

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

= \frac{1 \cdot 15}{u}

Multiply the numbers:

1 \cdot 15 = 15

= \frac{15}{u}

= 15 u + \frac{15}{u}

15 u + \frac{15}{u} = 34

Multiply both sides by

u

15 u + \frac{15}{u} = 34

Multiply both sides by

u

15 uu + \frac{15}{u} u = 34 u

Simplify

15 uu + \frac{15}{u} u = 34 u

Simplify

15 uu : 15 u^{2}

15 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 15 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 15 u^{2}

Simplify

\frac{15}{u} u : 15

\frac{15}{u} u

Multiply fractions:

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

= \frac{15 u}{u}

Cancel the common factor:

u

= 15

15 u^{2} + 15 = 34 u

15 u^{2} + 15 = 34 u

15 u^{2} + 15 = 34 u

Solve

15 u^{2} + 15 = 34 u : u = \frac{5}{3}, u = \frac{3}{5}

15 u^{2} + 15 = 34 u

Move

34 u

to the left side

15 u^{2} + 15 = 34 u

Subtract

34 u

from both sides

15 u^{2} + 15 - 34 u = 34 u - 34 u

Simplify

15 u^{2} + 15 - 34 u = 0

15 u^{2} + 15 - 34 u = 0

Write in the standard form

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

15 u^{2} - 34 u + 15 = 0

Solve with the quadratic formula

15 u^{2} - 34 u + 15 = 0

Quadratic Equation Formula:

For

a = 15, b = - 34, c = 15

u_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 15 \cdot 15}{2 \cdot 15}

u_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 15 \cdot 15}{2 \cdot 15}

(- 34)^{2} - 4 \cdot 15 \cdot 15 = 16

(- 34)^{2} - 4 \cdot 15 \cdot 15

Apply exponent rule:

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

n

is even

(- 34)^{2} = 3 4^{2}

= 3 4^{2} - 4 \cdot 15 \cdot 15

Multiply the numbers:

4 \cdot 15 \cdot 15 = 900

= 3 4^{2} - 900

3 4^{2} = 1156

= 1156 - 900

Subtract the numbers:

1156 - 900 = 256

= 256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 34 ) \pm 16}{2 \cdot 15}

Separate the solutions

u_{1} = \frac{- ( - 34 ) + 16}{2 \cdot 15}, u_{2} = \frac{- ( - 34 ) - 16}{2 \cdot 15}

u = \frac{- ( - 34 ) + 16}{2 \cdot 15} : \frac{5}{3}

\frac{- ( - 34 ) + 16}{2 \cdot 15}

Apply rule

- (- a) = a

= \frac{34 + 16}{2 \cdot 15}

Add the numbers:

34 + 16 = 50

= \frac{50}{2 \cdot 15}

Multiply the numbers:

2 \cdot 15 = 30

= \frac{50}{30}

Cancel the common factor:

10

= \frac{5}{3}

u = \frac{- ( - 34 ) - 16}{2 \cdot 15} : \frac{3}{5}

\frac{- ( - 34 ) - 16}{2 \cdot 15}

Apply rule

- (- a) = a

= \frac{34 - 16}{2 \cdot 15}

Subtract the numbers:

34 - 16 = 18

= \frac{18}{2 \cdot 15}

Multiply the numbers:

2 \cdot 15 = 30

= \frac{18}{30}

Cancel the common factor:

6

= \frac{3}{5}

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = \frac{3}{5}

u = \frac{5}{3}, u = \frac{3}{5}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(u + u^{- 1}) 15

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{5}{3}, u = \frac{3}{5}

u = \frac{5}{3}, u = \frac{3}{5}

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{5}{3} : x = ln (\frac{5}{3})

e^{x} = \frac{5}{3}

Apply exponent rules

e^{x} = \frac{5}{3}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{5}{3})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{5}{3})

x = ln (\frac{5}{3})

Solve

e^{x} = \frac{3}{5} : x = ln (\frac{3}{5})

e^{x} = \frac{3}{5}

Apply exponent rules

e^{x} = \frac{3}{5}

f (x) = g (x)

, then

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

ln (e^{x}) = ln (\frac{3}{5})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{3}{5})

x = ln (\frac{3}{5})

x = ln (\frac{5}{3}), x = ln (\frac{3}{5})

x = ln (\frac{5}{3}), x = ln (\frac{3}{5})

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Frequently Asked Questions (FAQ)

What is the general solution for cosh(x)= 17/15 ?
The general solution for cosh(x)= 17/15 is x=ln(5/3),x=ln(3/5)