Master the Huperbolic Equation: Solutions with Natural Logarithms

[fan_103]

1.Solve the equation sinh2x+ 4coshx = 6sinhx+ 12 giving your answers in terms of natural logarithms2. Sinh2x= 2SinhxCoshx
Cosh2x=2Cosh^2x-1
3. 2SinhxCoshx+4Coshx =6Sinhx+12
I don't know what to do next please help me!

 

[fan_103]

I ve got the answer! ;)

 

[Mark44]

Can you share it with us? I filled in most of a page last night, but stopped before getting solutions for x. There are two other identities that you didn't list: cosh x = (1/2)(e^x + e^-x) and sinh x = (1/2)(e^x - e^-x).

I converted the equation using these identities and eventually got to
e^2x - 2e^x - e^-2x - 10e^-x = 24.
I completed the square in the first pair of terms and the last pair of terms on the left side, and got to this equation:
(e^x - 1)² = (e^-x + 5)²
or
e^x - 1 = +/-(e^-x + 5)

It was getting past my bedtime, so I quit there.

After thinking about it a bit more this morning, I can see that the equation above can be split into two cases. Multiplying by e^x yields these two equations:
e^2x - 6e^x - 1 = 0
e^2x + 4e^x + 1 = 0

Each of these is a quadratic in e^x, so we can solve for e^x in each of them, and finally take the natural log to get x.

Is that about what you did?

 

[fan_103]

Can you share it with us? I filled in most of a page last night, but stopped before getting solutions for x. There are two other identities that you didn't list: cosh x = (1/2)(e^x + e^-x) and sinh x = (1/2)(e^x - e^-x).

I converted the equation using these identities and eventually got to
e^2x - 2e^x - e^-2x - 10e^-x = 24.
I completed the square in the first pair of terms and the last pair of terms on the left side, and got to this equation:
(e^x - 1)² = (e^-x + 5)²
or
e^x - 1 = +/-(e^-x + 5)

It was getting past my bedtime, so I quit there.

After thinking about it a bit more this morning, I can see that the equation above can be split into two cases. Multiplying by e^x yields these two equations:
e^2x - 6e^x - 1 = 0
e^2x + 4e^x + 1 = 0

Each of these is a quadratic in e^x, so we can solve for e^x in each of them, and finally take the natural log to get x.

Is that about what you did?

Its very easy indeed mate.No need to use $$e^-x$$
Sinh$$2x$$+4Cosh$$x$$=6Sinh$$x$$+12
2Sinh$$x$$Cosh$$x$$+4Cosh$$x$$-6Sinh$$x$$+12=0
2Cosh$$x$$(Sinh$$x$$+2)-6(Sinh$$x$$+2)=0
(2Cosh$$x$$-6)(Sinh$$x$$+2)=0

Either 2Cosh$$x$$-6=0
Coshx=3
$$x$$=Cosh^-1(3)
=ln(3+$$\sqrt{9-1}$$)
=ln(3+$$\sqrt{6}$$)
Or Sinh$$x$$=-2
$$x$$=Sinh^-1(-2)
=ln (-2+$$\sqrt{4+1}$$)
=ln($$\sqrt{5}$$)-2)
=ln(-2+$$\sqrt{5}$$)

 

[Mark44]

Its very easy indeed mate.No need to use $$e^-x$$
Sinh$$2x$$+4Cosh$$x$$=6Sinh$$x$$+12
2Sinh$$x$$Cosh$$x$$+4Cosh$$x$$-6Sinh$$x$$+12=0
2Cosh$$x$$(Sinh$$x$$+2)-6(Sinh$$x$$+2)=0
(2Cosh$$x$$-6)(Sinh$$x$$+2)=0

Either 2Cosh$$x$$-6=0
Coshx=3
$$x$$=Cosh^-1(3)
=ln(3+$$\sqrt{9-1}$$)
=ln(3+$$\sqrt{6}$$)
Or Sinh$$x$$=-2
$$x$$=Sinh^-1(-2)
=ln (-2+$$\sqrt{4+1}$$)
=ln($$\sqrt{5}$$)-2)
=ln(-2+$$\sqrt{5}$$)

Your way is simpler than mine, but after correcting a sign error I made, I get three solutions to your two. BTW, there is an error in the first solution you show. sqrt(9 - 1) = sqrt(8), not sqrt(6). So your two solutions are x = ln(3 + 2sqrt(2)) and x = ln(-2 + sqrt(5)).

My equation (e^x - 1)² = (e^-x + 5)² should have been (e^x - 1)² = (e^-x - 5)²

so that e^x - 1 = +/-(e^-x - 5)

This leads eventually to
e^2x - 6e^x + 1 = 0 and
e^2x + 4e^x - 1 = 0

Substituting u = e^x

We get u² -6u + 1 = 0, and
u² + 4u - 1 = 0

Using the quadratic formula on the first, we get
u = 3 +/- 2sqrt(2), so x = ln(3 +/- 2sqrt(2))

Using the quadratic formula on the second, we get
u = -2 + sqrt(5), so x = ln(-2 + sqrt(5))

The solution I have but you don't is x = ln(3 - 2sqrt(2)).

It's possible that your missing solution came from using cosh^-1, but I don't know this for sure.