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How do you isolate X in the following:


CrazCo
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[math]\ln (e^x + e^{-x}) = \ln 3[/math]

 

See what you can do using the properties of logarithms to tease x out of there.

There is no teasing that x out of there. Bignose suggestion to look into the hyperbolic functions is exactly right.

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My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0).

 

Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years.

 

(Hint: The answer to these is often = +/- .... likeyou have there)

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My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0).

 

Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years.

 

(Hint: The answer to these is often = +/- .... likeyou have there)

Yep. That will do, also, and that is probably what is wanted for this (what appears to be homework) problem.

 

Of course saying acosh(1.5) is much easier.

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The hint ed84c gave is a bit cryptic (and appropriately so). Several hours have passed.

 

A bit less cryptically, the idea is the solve for [math]e^x[/math]. Set [math]u=e^x[/math]. The equation in the original post becomes [math]u+1/u=3[/math].

 

Now multiply both sides by u. What kind of equation results from this?

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LIKE THIS?

 

e^x + 1/e^x = 3

 

e^2x -3(e^x)+1=0

 

Assume A = e^x

 

A^2 - 3A + 1 = 0

 

Quadratic Formula

 

3+- sqrt 9-4/2 = A

 

3+- sqrt5/2 = A

 

Replace

 

3+sqrt5/2 = e^x

 

x = ln(3+sqrt5/2) = .96

X = ln(3-sqrt5/2) = -.96

 

 

im proud i got it!!!!

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