How Do You Solve a Hyperbolic Function Using Natural Logs?

[KatieLynn]

Homework Statement

e^x - e^-x all divided by 2 = -1
express answer in natural logs

Homework Equations

No equations, just properties of logs and natural logs

The Attempt at a Solution

First, I multiplied everything by 2


e^x - e^-x = -2

then i took the natural log of both sides and brought down the exponent

(x)Ln(e) + (x)Ln(e) = -2

so you can simplify that i think to

2xLn(e) = -2

so x= -2/(2Lne)

which Lne is 1 so

x=-2/1

x=-2?

I don't think that's right because it says express answer in natural logs so it should simplify out that easily.

 

[sourlemon]

I think you missed something.

When you turn
e^x - e^-x = -2

into

(x)Ln(e) + (x)Ln(e) = -2

the -2 is part of it too. So it should be

(x)Ln(e) + (x)Ln(e) = Ln(-2)

 

[HallsofIvy]

Homework Statement

e^x - e^-x all divided by 2 = -1
express answer in natural logs

Homework Equations

No equations, just properties of logs and natural logs

The Attempt at a Solution

First, I multiplied everything by 2


e^x - e^-x = -2

then i took the natural log of both sides and brought down the exponent

(x)Ln(e) + (x)Ln(e) = -2

so you can simplify that i think to

2xLn(e) = -2

so x= -2/(2Lne)

which Lne is 1 so

x=-2/1

x=-2?

I don't think that's right because it says express answer in natural logs so it should simplify out that easily.

I think you missed something.

When you turn
e^x - e^-x = -2

into

(x)Ln(e) + (x)Ln(e) = -2

the -2 is part of it too. So it should be

(x)Ln(e) + (x)Ln(e) = Ln(-2)

Unfortunately you are both wrong. You can't just "turn e^x- e^-x= -2 into Ln(e^x)+ Ln(e^-x)".

You can't "take the natural logarithm of both sides because, as sourlemon said, you must also take the natural logarithm of -2 and that is not defined: the domain of Ln(x)is "all positive x". Even if it were +2 on the right side, Ln(x+ y) is NOT Ln(x)+ Ln(y).

Instead, go back to e^x- e^-x= -2 and multiply both sides by e^x: (e^x)^2- 1= 2(e^x) or y^2- 1= 2y (taking y= e^x) so y^2- 2y= 1. You can solve that by completing the square on the left: y^2- 2y+ 1= (y-1)^2= 2 so y= $1\pm\sqrt{2}$.
Once you have $e^x= 1+ \sqrt{2}$, then take the Ln of both sides. (Do you see why $e^x= 1- \sqrt{2}$ is not possible?

I beat you Compuchip!

 

[CompuChip]

Also note that in general, log(a + b) is not equal to log(a) + log(b). For example, log(1 + 0) = log(1) = 0 =/= log(1) + log(0) = $-\infty$. There is just a rule saying that log(a * b) = log(a) + log(b). Instead, multiply both sides by $e^x$ and first solve it for $y = e^x$ as a quadratic expression. Then you can take the logarithm.

[edit]Wow, you are fast [/edit]

 

[KatieLynn]

I think you can't do $e^x= 1- \sqrt{2}$ because it would be a negative answer and logs have to be greater than zero.I understand how you solved the problem but I was never taught to do it that way, are there other way to solve the problem other than completing the square?

Is it possible to factor it?