Solve for #x#: #164^(x+2) - 162^(x+1) + 1 = 0#?

Question

20342 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T17:04:36

#x=-4#

#16*4^(x+2) - 16*2^(x+1) + 1 = #
#=16^2cdot4^x-2cdot16cdot2^x+1=#
#=16^2cdot(2^x)^2-32cdot 2^x+1=0#

Making #y=2^x#

#16^2y^2-32y+1=0# solving for #y# we have

#y = 1/16=2^x=2^(-4)# then #x=-4#

Solve for #x#: #16*4^(x+2) - 16*2^(x+1) + 1 = 0#?