How do you solve #3^(2x+2) + 3^(x+1) - 12 = 0#?

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Answer 1 · 2015-08-16T22:39:48

Recognise as a quadratic in #3^(x+1)#, one of whose solutions gives us a Real solution, #x=0#.

Let #t = 3^(x+1)#

Then #0 = 3^(2x+2)+3^(x+1)-12 = t^2+t-12 = (t+4)(t-3)#

So #t = -4# or #t = 3#.

Now #3^(x+1) > 0# for all #x in RR#, so we can discard the case #t = -4#.

The remaining solution gives us:

#3^(x+1) = t = 3 = 3^1#

Since exponentiation is one-one as a function from #RR# to #(0, oo)#, this implies #x+1 = 1#, so #x = 0#.