What is the general solution for 2sinh(2x)-6cosh(x)=0 ?

The general solution for 2sinh(2x)-6cosh(x)=0 is x=ln(3.30277…)

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2sinh(2x)-6cosh(x)=0

Solution

2 sinh (2 x) - 6 cosh (x) = 0

Solution

x = ln (3.30277 \dots)

Degrees

x = 68.45488 \dots^{\circ}

Solution steps

2 sinh (2 x) - 6 cosh (x) = 0

Rewrite using trig identities

2 sinh (2 x) - 6 cosh (x) = 0

Use the Hyperbolic identity:

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

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 cosh (x) = 0

Use the Hyperbolic identity:

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

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 \cdot \frac{e ^{x} + e ^{- x}}{2} = 0

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 \cdot \frac{e ^{x} + e ^{- x}}{2} = 0

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 \cdot \frac{e ^{x} + e ^{- x}}{2} = 0 : x = ln (3.30277 \dots)

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 \cdot \frac{e ^{x} + e ^{- x}}{2} = 0

Add

6 \frac{e ^{x} + e ^{- x}}{2}

to both sides

2 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} - 6 \cdot \frac{e ^{x} + e ^{- x}}{2} + 6 \cdot \frac{e ^{x} + e ^{- x}}{2} = 0 + 6 \cdot \frac{e ^{x} + e ^{- x}}{2}

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}, e^{- x} = (e^{x})^{- 1}

(e^{x})^{2} - (e^{x})^{- 2} = 3 (e^{x} + (e^{x})^{- 1})

(e^{x})^{2} - (e^{x})^{- 2} = 3 (e^{x} + (e^{x})^{- 1})

Rewrite the equation with

e^{x} = u

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

Solve

u^{2} - u^{- 2} = 3 (u + u^{- 1}) : u \approx - 0.30277 \dots, u \approx 3.30277 \dots

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

Refine

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

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

Simplify

u^{2} u^{2} : u^{4}

u^{2} u^{2}

Apply exponent rule:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

Simplify

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

- \frac{1}{u ^{2}} u^{2}

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= - 1

u^{4} - 1 = 3 (u + \frac{1}{u}) u^{2}

u^{4} - 1 = 3 (u + \frac{1}{u}) u^{2}

u^{4} - 1 = 3 (u + \frac{1}{u}) u^{2}

Expand

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

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

= 3 u^{2} u + 3 u^{2} \frac{1}{u}

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

Simplify

3 u^{2} u + 3 \cdot \frac{1}{u} u^{2} : 3 u^{3} + 3 u

3 u^{2} u + 3 \cdot \frac{1}{u} u^{2}

3 u^{2} u = 3 u^{3}

3 u^{2} u

Apply exponent rule:

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

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

= 3 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 3 u^{3}

3 \cdot \frac{1}{u} u^{2} = 3 u

3 \cdot \frac{1}{u} u^{2}

Multiply fractions:

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

= \frac{1 \cdot 3 u ^{2}}{u}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 u ^{2}}{u}

Cancel the common factor:

u

= 3 u

= 3 u^{3} + 3 u

= 3 u^{3} + 3 u

u^{4} - 1 = 3 u^{3} + 3 u

Solve

u^{4} - 1 = 3 u^{3} + 3 u : u \approx - 0.30277 \dots, u \approx 3.30277 \dots

u^{4} - 1 = 3 u^{3} + 3 u

Move

3 u

to the left side

u^{4} - 1 = 3 u^{3} + 3 u

Subtract

3 u

from both sides

u^{4} - 1 - 3 u = 3 u^{3} + 3 u - 3 u

Simplify

u^{4} - 1 - 3 u = 3 u^{3}

u^{4} - 1 - 3 u = 3 u^{3}

Move

3 u^{3}

to the left side

u^{4} - 1 - 3 u = 3 u^{3}

Subtract

3 u^{3}

from both sides

u^{4} - 1 - 3 u - 3 u^{3} = 3 u^{3} - 3 u^{3}

Simplify

u^{4} - 1 - 3 u - 3 u^{3} = 0

u^{4} - 1 - 3 u - 3 u^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 3 u^{3} - 3 u - 1 = 0

Find one solution for

u^{4} - 3 u^{3} - 3 u - 1 = 0

using Newton-Raphson:

u \approx - 0.30277 \dots

u^{4} - 3 u^{3} - 3 u - 1 = 0

Newton-Raphson Approximation Definition

f (u) = u^{4} - 3 u^{3} - 3 u - 1

Find

f^{^{'}} (u) : 4 u^{3} - 9 u^{2} - 3

\frac{d}{d u} (u^{4} - 3 u^{3} - 3 u - 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{4}) - \frac{d}{d u} (3 u^{3}) - \frac{d}{d u} (3 u) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 u^{4 - 1}

Simplify

= 4 u^{3}

\frac{d}{d u} (3 u^{3}) = 9 u^{2}

\frac{d}{d u} (3 u^{3})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

= 9 u^{2}

\frac{d}{d u} (3 u) = 3

\frac{d}{d u} (3 u)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 3 \cdot 1

Simplify

= 3

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 4 u^{3} - 9 u^{2} - 3 - 0

Simplify

= 4 u^{3} - 9 u^{2} - 3

Let

u_{0} = 0

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 0.33333 \dots : Δ u_{1} = 0.33333 \dots

f (u_{0}) = 0^{4} - 3 \cdot 0^{3} - 3 \cdot 0 - 1 = - 1

f^{^{'}} (u_{0}) = 4 \cdot 0^{3} - 9 \cdot 0^{2} - 3 = - 3

u_{1} = - 0.33333 \dots

Δ u_{1} = ∣ - 0.33333 \dots - 0 ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = - 0.30357 \dots : Δ u_{2} = 0.02976 \dots

f (u_{1}) = (- 0.33333 \dots)^{4} - 3 (- 0.33333 \dots)^{3} - 3 (- 0.33333 \dots) - 1 = 0.12345 \dots

f^{^{'}} (u_{1}) = 4 (- 0.33333 \dots)^{3} - 9 (- 0.33333 \dots)^{2} - 3 = - 4.14814 \dots

u_{2} = - 0.30357 \dots

Δ u_{2} = ∣ - 0.30357 \dots - (- 0.33333 \dots) ∣ = 0.02976 \dots

Δ u_{2} = 0.02976 \dots

u_{3} = - 0.30277 \dots : Δ u_{3} = 0.00079 \dots

f (u_{2}) = (- 0.30357 \dots)^{4} - 3 (- 0.30357 \dots)^{3} - 3 (- 0.30357 \dots) - 1 = 0.00313 \dots

f^{^{'}} (u_{2}) = 4 (- 0.30357 \dots)^{3} - 9 (- 0.30357 \dots)^{2} - 3 = - 3.94130 \dots

u_{3} = - 0.30277 \dots

Δ u_{3} = ∣ - 0.30277 \dots - (- 0.30357 \dots) ∣ = 0.00079 \dots

Δ u_{3} = 0.00079 \dots

u_{4} = - 0.30277 \dots : Δ u_{4} = 5.27302 E - 7

f (u_{3}) = (- 0.30277 \dots)^{4} - 3 (- 0.30277 \dots)^{3} - 3 (- 0.30277 \dots) - 1 = 2.07551 E - 6

f^{^{'}} (u_{3}) = 4 (- 0.30277 \dots)^{3} - 9 (- 0.30277 \dots)^{2} - 3 = - 3.93608 \dots

u_{4} = - 0.30277 \dots

Δ u_{4} = ∣ - 0.30277 \dots - (- 0.30277 \dots) ∣ = 5.27302 E - 7

Δ u_{4} = 5.27302 E - 7

u \approx - 0.30277 \dots

Apply long division:

\frac{u ^{4} - 3 u ^{3} - 3 u - 1}{u + 0.30277 \dots} = u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots

u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots \approx 0

Find one solution for

u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots = 0

using Newton-Raphson:

u \approx 3.30277 \dots

u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots = 0

Newton-Raphson Approximation Definition

f (u) = u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots

Find

f^{^{'}} (u) : 3 u^{2} - 6.60555 \dots u + 1.00000 \dots

\frac{d}{d u} (u^{3} - 3.30277 \dots u^{2} + u - 3.30277 \dots)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (3.30277 \dots u^{2}) + \frac{d u}{d u} - \frac{d}{d u} (3.30277 \dots)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Simplify

= 3 u^{2}

\frac{d}{d u} (3.30277 \dots u^{2}) = 6.60555 \dots u

\frac{d}{d u} (3.30277 \dots u^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3.30277 \dots \frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3.30277 \dots \cdot 2 u^{2 - 1}

Simplify

= 6.60555 \dots u

\frac{d u}{d u} = 1

\frac{d u}{d u}

Apply the common derivative:

\frac{d u}{d u} = 1

= 1

\frac{d}{d u} (3.30277 \dots) = 0

\frac{d}{d u} (3.30277 \dots)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} - 6.60555 \dots u + 1 - 0

Simplify

= 3 u^{2} - 6.60555 \dots u + 1.00000 \dots

Let

u_{0} = 3

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = 3.36999 \dots : Δ u_{1} = 0.36999 \dots

f (u_{0}) = 3^{3} - 3.30277 \dots \cdot 3^{2} + 3 - 3.30277 \dots = - 3.02775 \dots

f^{^{'}} (u_{0}) = 3 \cdot 3^{2} - 6.60555 \dots \cdot 3 + 1.00000 \dots = 8.18334 \dots

u_{1} = 3.36999 \dots

Δ u_{1} = ∣ 3.36999 \dots - 3 ∣ = 0.36999 \dots

Δ u_{1} = 0.36999 \dots

u_{2} = 3.30515 \dots : Δ u_{2} = 0.06483 \dots

f (u_{1}) = 3.36999 \dots^{3} - 3.30277 \dots \cdot 3.36999 \dots^{2} + 3.36999 \dots - 3.30277 \dots = 0.83055 \dots

f^{^{'}} (u_{1}) = 3 \cdot 3.36999 \dots^{2} - 6.60555 \dots \cdot 3.36999 \dots + 1.00000 \dots = 12.80985 \dots

u_{2} = 3.30515 \dots

Δ u_{2} = ∣ 3.30515 \dots - 3.36999 \dots ∣ = 0.06483 \dots

Δ u_{2} = 0.06483 \dots

u_{3} = 3.30277 \dots : Δ u_{3} = 0.00237 \dots

f (u_{2}) = 3.30515 \dots^{3} - 3.30277 \dots \cdot 3.30515 \dots^{2} + 3.30515 \dots - 3.30277 \dots = 0.02834 \dots

f^{^{'}} (u_{2}) = 3 \cdot 3.30515 \dots^{2} - 6.60555 \dots \cdot 3.30515 \dots + 1.00000 \dots = 11.93974 \dots

u_{3} = 3.30277 \dots

Δ u_{3} = ∣ 3.30277 \dots - 3.30515 \dots ∣ = 0.00237 \dots

Δ u_{3} = 0.00237 \dots

u_{4} = 3.30277 \dots : Δ u_{4} = 3.12826 E - 6

f (u_{3}) = 3.30277 \dots^{3} - 3.30277 \dots \cdot 3.30277 \dots^{2} + 3.30277 \dots - 3.30277 \dots = 0.00003 \dots

f^{^{'}} (u_{3}) = 3 \cdot 3.30277 \dots^{2} - 6.60555 \dots \cdot 3.30277 \dots + 1.00000 \dots = 11.90836 \dots

u_{4} = 3.30277 \dots

Δ u_{4} = ∣ 3.30277 \dots - 3.30277 \dots ∣ = 3.12826 E - 6

Δ u_{4} = 3.12826 E - 6

u_{5} = 3.30277 \dots : Δ u_{5} = 5.42823 E - 12

f (u_{4}) = 3.30277 \dots^{3} - 3.30277 \dots \cdot 3.30277 \dots^{2} + 3.30277 \dots - 3.30277 \dots = 6.46412 E - 11

f^{^{'}} (u_{4}) = 3 \cdot 3.30277 \dots^{2} - 6.60555 \dots \cdot 3.30277 \dots + 1.00000 \dots = 11.90832 \dots

u_{5} = 3.30277 \dots

Δ u_{5} = ∣ 3.30277 \dots - 3.30277 \dots ∣ = 5.42823 E - 12

Δ u_{5} = 5.42823 E - 12

u \approx 3.30277 \dots

Apply long division:

\frac{u ^{3} - 3.30277 \dots u ^{2} + u - 3.30277 \dots}{u - 3.30277 \dots} = u^{2} + 1

u^{2} + 1 \approx 0

Find one solution for

u^{2} + 1 = 0

using Newton-Raphson:

No Solution for

u \in R

u^{2} + 1 = 0

Newton-Raphson Approximation Definition

f (u) = u^{2} + 1

Find

f^{^{'}} (u) : 2 u

\frac{d}{d u} (u^{2} + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplify

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

= 2 u + 0

Simplify

= 2 u

Let

u_{0} = - 2

Compute

u_{n + 1}

until

Δ u_{n + 1} < 0.000001

u_{1} = - 0.74999 \dots : Δ u_{1} = 1.25

f (u_{0}) = (- 2)^{2} + 1 = 5.00000 \dots

f^{^{'}} (u_{0}) = 2 (- 2) = - 4.00000 \dots

u_{1} = - 0.74999 \dots

Δ u_{1} = ∣ - 0.74999 \dots - (- 2) ∣ = 1.25

Δ u_{1} = 1.25

u_{2} = 0.29166 \dots : Δ u_{2} = 1.04166 \dots

f (u_{1}) = (- 0.74999 \dots)^{2} + 1 = 1.56250 \dots

f^{^{'}} (u_{1}) = 2 (- 0.74999 \dots) = - 1.5

u_{2} = 0.29166 \dots

Δ u_{2} = ∣ 0.29166 \dots - (- 0.74999 \dots) ∣ = 1.04166 \dots

Δ u_{2} = 1.04166 \dots

u_{3} = - 1.56845 \dots : Δ u_{3} = 1.86011 \dots

f (u_{2}) = 0.29166 \dots^{2} + 1 = 1.08506 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.29166 \dots = 0.58333 \dots

u_{3} = - 1.56845 \dots

Δ u_{3} = ∣ - 1.56845 \dots - 0.29166 \dots ∣ = 1.86011 \dots

Δ u_{3} = 1.86011 \dots

u_{4} = - 0.46544 \dots : Δ u_{4} = 1.10301 \dots

f (u_{3}) = (- 1.56845 \dots)^{2} + 1 = 3.46004 \dots

f^{^{'}} (u_{3}) = 2 (- 1.56845 \dots) = - 3.13690 \dots

u_{4} = - 0.46544 \dots

Δ u_{4} = ∣ - 0.46544 \dots - (- 1.56845 \dots) ∣ = 1.10301 \dots

Δ u_{4} = 1.10301 \dots

u_{5} = 0.84153 \dots : Δ u_{5} = 1.30697 \dots

f (u_{4}) = (- 0.46544 \dots)^{2} + 1 = 1.21663 \dots

f^{^{'}} (u_{4}) = 2 (- 0.46544 \dots) = - 0.93088 \dots

u_{5} = 0.84153 \dots

Δ u_{5} = ∣ 0.84153 \dots - (- 0.46544 \dots) ∣ = 1.30697 \dots

Δ u_{5} = 1.30697 \dots

u_{6} = - 0.17339 \dots : Δ u_{6} = 1.01492 \dots

f (u_{5}) = 0.84153 \dots^{2} + 1 = 1.70817 \dots

f^{^{'}} (u_{5}) = 2 \cdot 0.84153 \dots = 1.68306 \dots

u_{6} = - 0.17339 \dots

Δ u_{6} = ∣ - 0.17339 \dots - 0.84153 \dots ∣ = 1.01492 \dots

Δ u_{6} = 1.01492 \dots

u_{7} = 2.79697 \dots : Δ u_{7} = 2.97036 \dots

f (u_{6}) = (- 0.17339 \dots)^{2} + 1 = 1.03006 \dots

f^{^{'}} (u_{6}) = 2 (- 0.17339 \dots) = - 0.34678 \dots

u_{7} = 2.79697 \dots

Δ u_{7} = ∣ 2.79697 \dots - (- 0.17339 \dots) ∣ = 2.97036 \dots

Δ u_{7} = 2.97036 \dots

u_{8} = 1.21972 \dots : Δ u_{8} = 1.57725 \dots

f (u_{7}) = 2.79697 \dots^{2} + 1 = 8.82306 \dots

f^{^{'}} (u_{7}) = 2 \cdot 2.79697 \dots = 5.59394 \dots

u_{8} = 1.21972 \dots

Δ u_{8} = ∣ 1.21972 \dots - 2.79697 \dots ∣ = 1.57725 \dots

Δ u_{8} = 1.57725 \dots

u_{9} = 0.19993 \dots : Δ u_{9} = 1.01979 \dots

f (u_{8}) = 1.21972 \dots^{2} + 1 = 2.48772 \dots

f^{^{'}} (u_{8}) = 2 \cdot 1.21972 \dots = 2.43944 \dots

u_{9} = 0.19993 \dots

Δ u_{9} = ∣ 0.19993 \dots - 1.21972 \dots ∣ = 1.01979 \dots

Δ u_{9} = 1.01979 \dots

u_{10} = - 2.40088 \dots : Δ u_{10} = 2.60081 \dots

f (u_{9}) = 0.19993 \dots^{2} + 1 = 1.03997 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.19993 \dots = 0.39986 \dots

u_{10} = - 2.40088 \dots

Δ u_{10} = ∣ - 2.40088 \dots - 0.19993 \dots ∣ = 2.60081 \dots

Δ u_{10} = 2.60081 \dots

Cannot find solution

The solutions are

u \approx - 0.30277 \dots, u \approx 3.30277 \dots

u \approx - 0.30277 \dots, u \approx 3.30277 \dots

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u^{2} - u^{- 2}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Take the denominator(s) of

3 (u + u^{- 1})

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u \approx - 0.30277 \dots, u \approx 3.30277 \dots

u \approx - 0.30277 \dots, u \approx 3.30277 \dots

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = - 0.30277 \dots :

No Solution for

x \in R

e^{x} = - 0.30277 \dots

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

e^{x} = 3.30277 \dots : x = ln (3.30277 \dots)

e^{x} = 3.30277 \dots

Apply exponent rules

e^{x} = 3.30277 \dots

f (x) = g (x)

, then

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

ln (e^{x}) = ln (3.30277 \dots)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (3.30277 \dots)

x = ln (3.30277 \dots)

x = ln (3.30277 \dots)

x = ln (3.30277 \dots)

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Frequently Asked Questions (FAQ)

What is the general solution for 2sinh(2x)-6cosh(x)=0 ?
The general solution for 2sinh(2x)-6cosh(x)=0 is x=ln(3.30277…)