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Answer 1 · 2018-01-28T06:53:10

#x = log_3 2#

#9^x - 3^x - 2 = 0#

#rArr (3^2)^x - 3^x - 2 = 0#

#rArr (3^x)^2 - 3^x - 2 = 0#

Let #3^x = a#

So, #a^2 - a - 2 = 0#

#rArr a^2 + a - 2a - 2 = 0#

#rArr a(a + 1) - 2(a + 1) = 0#

#rArr (a + 1)(a - 2) = 0#

We get, #a + 1 = 0 rArr a = -1 rArr 3^x = -1#

So, No real value for x here.

And #a - 2 = 0 rArr a = 2 rArr 3^x = 2 rArr x = log_3 2#

So, #x = log_3 2#

Answer 2 · 2018-01-28T06:54:32

The answer is #x=log_3(2)#.

#9^x-3^x-2=0#

#(3^2)^x-3^x-2=0#

#3^(2x)-3^x-2=0#

#(3^x)^2-3^x-2=0#

Let #u = 3^x#:

#u^2-u-2=0#

#(u-2)(u+1)=0#

#u=-1,2#

Replace #u# back with #3^x#:

#3^x=2#

#cancel(3^x=-1)# - no solution

#log_cancel(3)(cancel(3)^x)=log_3(2)#

#x=log_3(2)~~0.6309#