2e^-x-e^x=0

Question

solve the equation 

Answer 1

2e^-x - e^x = 0 [add e^x to both sides]

2e^-x = e^x [divide e^-x on both sides]

2 = e^x/e^-x [use exponent quotient rule-subtract exponents]

2 = e^{(x - (-x))}[add exponents]

2= e^2x [take the natural logarithm of both sides]

ln(2) = ln(e^2x)[use logarithm exponent rule]

ln(2) = 2x*ln(e)[ln(e) = 1]

ln(2) = 2x [multiply both sides by 1/2]

 

x = (1/2)*ln(2)

 

 

 

 

Answer 2

    2e^-x - e^x  = 0

 

   factor out e^x

 

e^x ( 2e^-2x - 1 )  = 0

 

set both factors = 0 and solve for x

 

e^x = 0  ;  e^x will never equal zero, no solution here

 

2e^-2x - 1 = 0

 

       2e^-2x  = 1

 

         e^-2x  = 1/2

 

change from exponential form to logarithmic form

 

          -2x = log_e (1/2)

 

          -2x = ln (2^-1)

 

             x = -(1/2) ln (2^-1)

 

             x = (1/2) ln 2

 

There are several ways that the final answer can be written,

all of these ways could be changed to look the same

Answer 3

Let y = e^x.  Note that e^-x = 1/y

Then, 2/y - y = 0

2/y = y

y^2 = 2

y = +- sqrt(2)

Recall that y = e^x.

If x is real, then y can't be negative, so only sqrt(2) is a solution.

sqrt(2) = e^x

Taking the ln of both sides,

x = ln(sqrt(2)) = ln(2^(1/2)) = 1/2 ln(2)