Exponential equations

Question

e^x+e^-x=6

 

Answer 1

OK. We start with e^x+e^-x=6.

 

I multiplied through by e^x. This led to

 

e^2x - 6e^x + 1 = 0

 

Then I decided to set y = e^x. So now it's

 

y² - 6y + 1 = 0

 

This you solve with the Quadratic Formula and get

 

y = (6 ± √(36 - 4))/2 or y = 5.8 or 0.2 approx.

 

So e^x = 5.8 or 0.2. This means x = 1.76 or -1.61.

 

Sanity check: I ran this problem through Wolfram Alpha and got ±1.76. This is why you run sanity checks. No doubt I made a rounding error on the negative value. I apologize for that.

 

Cheers!

 

 

Answer 2

To see the equation better: First let e^x be any variable that you like, for example: m = e^x

 

That will the equation:

 

m + 1/m = 6

 

Multiplying both sides by 'm':

 

m² + 1 = 6m

 

Placing everything in one side:

 

m² - 6m + 1 = 0

 

Using quadratic formula:

 

m = [-(-6) ± √((-6)² - 4(1)(1))] / 2(1)

 

m = 3 ± 2√2

 

But m = e^x

 

e^x = 3 + 2√2  and  e^x = 3 - 2√2 *Both are positive values so there will be no erroneous solution*

 

x = ln(3 + 2√2) and  x = ln(3 - 2√2)