1)2^x+2^x+3=36->
2^x+2^x*2^3=36
2^x+8*2^x=36
9*2^x=36 |(:9)
2^x=4
->x=2
2)49^x=6*7^x+7
7^2^x=6*7^x+7
7^2x-6*7^x+7=0
7^x=t , t>=0
t^2-6t-7=0
Δ=36+28
Δ=64
t1=6+8/2
t1=7
t2=6-8/2
t2=-2/2 =-1 ..t = -1 . Nu respecta conditiile ,astfel doar t1=7 Este solutie
7^x=t1 -->>>
7^x=7 ->>x=1
