a)inpartim ambii termeni cu 3^x
3*2^x/3^x=2*3^3/3^x
3*(2/3)^x=2
(2/3)^x=2/3⇒x=1
b l/am rezolvat mai devreme .are sol unica x=0.
c)inpartim toti termenii ec cu 5^2x
2*5^2x/5^2x=2^x*5^x/5^2x+2^2x/5^2x
2-(2/5)^x-(2/5)^2x=0
notam 2/5 la put x cu t
t^2+t-2=0
rezolvam ec de gr 2 obtinem delta 9
x1=(-1+3)/2=1
X2=(-1-3)/2=-2
dar t>0 deci ( 2/5)^x=1⇒(2/5)^x=(2/5)^0⇒x=0
