What is the general solution for 5cosh(x)-3sinh(x)=5 ?

The general solution for 5cosh(x)-3sinh(x)=5 is x=2ln(2),x=0

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5cosh(x)-3sinh(x)=5

Solution

5 cosh (x) - 3 sinh (x) = 5

Solution

x = 2 ln (2), x = 0

Degrees

x = 79.42881 \dots^{\circ}, x = 0^{\circ}

Solution steps

5 cosh (x) - 3 sinh (x) = 5

Rewrite using trig identities

5 cosh (x) - 3 sinh (x) = 5

Use the Hyperbolic identity:

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

5 cosh (x) - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Use the Hyperbolic identity:

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

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5 : x = 2 ln (2), x = 0

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Apply exponent rules

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

Apply exponent rule:

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

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

5 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} - 3 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

5 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} - 3 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

Rewrite the equation with

e^{x} = u

5 \cdot \frac{u + ( u ) ^{- 1}}{2} - 3 \cdot \frac{u - ( u ) ^{- 1}}{2} = 5

Solve

5 \cdot \frac{u + u ^{- 1}}{2} - 3 \cdot \frac{u - u ^{- 1}}{2} = 5 : u = 4, u = 1

5 \cdot \frac{u + u ^{- 1}}{2} - 3 \cdot \frac{u - u ^{- 1}}{2} = 5

Refine

\frac{5 ( u ^{2} + 1 )}{2 u} - \frac{3 ( u ^{2} - 1 )}{2 u} = 5

Multiply both sides by

2 u

\frac{5 ( u ^{2} + 1 )}{2 u} - \frac{3 ( u ^{2} - 1 )}{2 u} = 5

Multiply both sides by

2 u

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u - \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u = 5 \cdot 2 u

Simplify

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u - \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u = 5 \cdot 2 u

Simplify

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u : 5 (u^{2} + 1)

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u

Multiply fractions:

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

= \frac{5 ( u ^{2} + 1 ) \cdot 2 u}{2 u}

Cancel the common factor:

2

= \frac{5 ( u ^{2} + 1 ) u}{u}

Cancel the common factor:

u

= 5 (u^{2} + 1)

Simplify

- \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u : - 3 (u^{2} - 1)

- \frac{3 ( u ^{2} - 1 )}{2 u} \cdot 2 u

Multiply fractions:

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

= - \frac{3 ( u ^{2} - 1 ) \cdot 2 u}{2 u}

Cancel the common factor:

2

= - \frac{3 ( u ^{2} - 1 ) u}{u}

Cancel the common factor:

u

= - 3 (u^{2} - 1)

Simplify

5 \cdot 2 u : 10 u

5 \cdot 2 u

Multiply the numbers:

5 \cdot 2 = 10

= 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

Solve

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u : u = 4, u = 1

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

Expand

5 (u^{2} + 1) - 3 (u^{2} - 1) : 2 u^{2} + 8

5 (u^{2} + 1) - 3 (u^{2} - 1)

Expand

5 (u^{2} + 1) : 5 u^{2} + 5

5 (u^{2} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 5, b = u^{2}, c = 1

= 5 u^{2} + 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 u^{2} + 5

= 5 u^{2} + 5 - 3 (u^{2} - 1)

Expand

- 3 (u^{2} - 1) : - 3 u^{2} + 3

- 3 (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = u^{2}, c = 1

= - 3 u^{2} - (- 3) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 3 u^{2} + 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= - 3 u^{2} + 3

= 5 u^{2} + 5 - 3 u^{2} + 3

Simplify

5 u^{2} + 5 - 3 u^{2} + 3 : 2 u^{2} + 8

5 u^{2} + 5 - 3 u^{2} + 3

Group like terms

= 5 u^{2} - 3 u^{2} + 5 + 3

Add similar elements:

5 u^{2} - 3 u^{2} = 2 u^{2}

= 2 u^{2} + 5 + 3

Add the numbers:

5 + 3 = 8

= 2 u^{2} + 8

= 2 u^{2} + 8

2 u^{2} + 8 = 10 u

Move

10 u

to the left side

2 u^{2} + 8 = 10 u

Subtract

10 u

from both sides

2 u^{2} + 8 - 10 u = 10 u - 10 u

Simplify

2 u^{2} + 8 - 10 u = 0

2 u^{2} + 8 - 10 u = 0

Write in the standard form

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

2 u^{2} - 10 u + 8 = 0

Solve with the quadratic formula

2 u^{2} - 10 u + 8 = 0

Quadratic Equation Formula:

For

a = 2, b = - 10, c = 8

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 \cdot 8 = 6

(- 10)^{2} - 4 \cdot 2 \cdot 8

Apply exponent rule:

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

n

is even

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

= 1 0^{2} - 4 \cdot 2 \cdot 8

Multiply the numbers:

4 \cdot 2 \cdot 8 = 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:

n a^{n} = a

6^{2} = 6

= 6

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

10 + 6 = 16

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= 4

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

10 - 6 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 4, u = 1

u = 4, u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

5 \frac{u + u ^{- 1}}{2} - 3 \frac{u - u ^{- 1}}{2}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 4, u = 1

u = 4, u = 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 4 : x = 2 ln (2)

e^{x} = 4

Apply exponent rules

e^{x} = 4

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (4)

Simplify

ln (4) : 2 ln (2)

ln (4)

Rewrite

4

in power-base form:

4 = 2^{2}

= ln (2^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= 2 ln (2)

x = 2 ln (2)

x = 2 ln (2)

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

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

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

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

x = 2 ln (2), x = 0

x = 2 ln (2), x = 0

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5cosh(x)-3sinh(x)=5 ?
The general solution for 5cosh(x)-3sinh(x)=5 is x=2ln(2),x=0