Laura Cristina 08/10/2023 at 6:30 pm 4^x+8^x=16^x | : 4^x 1+2^x-4^x=0 2^x= t t²-t-1 =0 t1,2= (1+/-sqrt5)/2 2^x = 1/2+/- sqrt5/2 x= [log (sqrt5/2+1/2)/log 2] + [(2i pi n1/ log 2] x= [ log (sqrt5/2-1/2)/ log 2]+ [i pi(2n2+1)/ log2] n1,n2 ( belongs Z) Reply
4^x+8^x=16^x | : 4^x
1+2^x-4^x=0
2^x= t t²-t-1 =0 t1,2= (1+/-sqrt5)/2
2^x = 1/2+/- sqrt5/2
x= [log (sqrt5/2+1/2)/log 2] +
[(2i pi n1/ log 2]
x= [ log (sqrt5/2-1/2)/ log 2]+ [i pi(2n2+1)/ log2] n1,n2 ( belongs Z)