Solving for 'x' in difficult equation

[Log0008]

Hello all,

I am just interested into how you solve the following for x

 
 

and it is g'(x) =-1

[grannysmith]

e^(-x/2)=e^(x/2) * e^(-1)

[Log0008]

That has confused me mor, my understanding was you had to let e^x/2=y then use quadratic formula (based on hint from teacher)

[Stevensmay]

That has confused me mor, my understanding was you had to let e^x/2=y then use quadratic formula (based on hint from teacher)

Multiply everything in your equation by e^(x/2), then rewrite it with y = e^(x/2).

Should end up with y^2 -2y - 1 = 0

[fred42]

Please see attached.