Japan | A Nice Math Olympiad Exponential Problem | 2^a+2^-a=4 | Find a=? | How to solve?

[@adrianodeamarante238]

log(a*b)=log(a)+log(b)≠ log(a+b) Check please!

[@mircoceccarelli6689]

👍 t = 2^a => t^2 - 4t + 1 = 0 t^2 - 4t + 4 = 3 ( t - 2 )^2 = 3 ... t = 2 +/- sqrt( 3 ) > 0 2^a = 2 +/- sqrt( 3 ) a = [ ln( 2 +/- sqrt( 3 ) ) ]/ln( 2 ) 😊👍👋