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

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Answer 1 · 2015-08-16T22:39:48

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

Let $t = 3^{x + 1}$ $t = 3^(x+1)$

Then $0 = 3^{2 x + 2} + 3^{x + 1} - 12 = t^{2} + t - 12 = (t + 4) (t - 3)$ $0 = 3^(2x+2)+3^(x+1)-12 = t^2+t-12 = (t+4)(t-3)$

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

Now $3^{x + 1} > 0$ $3^(x+1) > 0$ 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 = 032x+2+3x+1−12=03^(2x+2) + 3^(x+1) - 12 = 0?