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 $x = 0$ .

How do you solve 3^(2x+2) + 3^(x+1) - 12 = 03^(2x+2) + 3^(x+1) - 12 = 0?