Question Number 104475 by pete last updated on 21/Jul/20

Solve 2^(2x)  + 42^x −32=0

Answered by $@y@m last updated on 21/Jul/20

I think there is a typo in question.  It should be:  Solve 2^(2x)  + 4.2^x −32=0  Solution: Let 2^x =y  y^2 +4y−32=0  y=4,−8  But y=−8 is inadmissible.  ∴ y=4  x=2

Commented bypete last updated on 21/Jul/20

ok thanks very much sir

Answered by Dwaipayan Shikari last updated on 21/Jul/20

2^(2x) +4.2^x −32=0  p^2 +4p−32=0   {2^x =p  p=−8,4  2^x =4⇒x=2  2^x =−8  xlog2=log(−1)+log(8)  x=((log(e^(πi) )+log(8))/(log(2)))=((πi+log8)/(log2))(Another solution)

Commented by$@y@m last updated on 22/Jul/20