How do you solve $(\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}$ ?

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Answer 1 · 2018-01-20T10:27:03

$x=-4$ or $x=-2$

Start by working on the left side:

Since $81=9^2$ we know $1/81=1/9^2$ and by the rules $1/a=a^-1$ , we know that $1/9^2 = 9^-2$

By the rule $(a^b)^c=a^(bc)$ , we know

$(9^-2)^(6x+2) =9^(-12x-4)$ .

So our equation now looks like:

$9^(-12x-4)=9^(2x^2+12)$

If $a^b = a^c$ , then $b=c$ , so:

$-12x-4 = 2x^2+12$

moving everything to one side and collecting like terms we have:

$2x^2+12x+16=0$

Dividing thruogh by 2:

$x^2+6x+8=0$

factoring:

$(x+4)(x+2)=0$

Solving:

$x=-4$ or $x=-2$

How do you solve (\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}(\frac { 1} { 81} ) ^ { 6x + 2} = 9^ { 2x ^ { 2} + 12}?