$x$ in the Exponents? Not again!

We have $\frac{8^x+27^x}{12^x+18^x} \cdot \frac{\frac{1}{27^x}}{\frac{1}{27^x}}=\frac{7}{6}$ Which yields $\frac{ \left[\left(\frac{2}{3} \right)^x\right]^3 +1 }{ \left[\left(\frac{2}{3} \right)^x\right]^2 +\left(\frac{2}{3} \right)^x}=\frac{7}{6}$ Now, let $\displaystyle y=\left(\frac{2}{3} \right)^x$ , then we have $\frac{y^3+1}{y^2+y}=\frac{7}{6}$ which can be simplified into $y+\frac{1}{y}-1=\frac{7}{6}$ Solving for $y$ gives $\displaystyle y=\frac{2}{3}, \frac{3}{2}$ And since $\displaystyle y=\left(\frac{2}{3} \right)^x$ , we have $\displaystyle \left(\frac{2}{3} \right)^x=\frac{2}{3}, \frac{3}{2}$ which implies that $x=\pm 1$ Thus, $\sqrt{2a^2+2b^2}=\sqrt{2+2}=\boxed{2}$

Since $8^x+27^x=(2^x)^3+(3^x)^3=(2^x+3^x)((2^x)^2-(2^x)(3^x)+(3^x)^2)=(2^x+3^x)(4^x-6^x+9^x)$ by factorization of the sum of two cubes, and $12^x+18^x=6^x(2^x+3^x)$ by factoring by GCF, then simplication of the fraction yields $\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x+3^x)(4^x-6^x+9^x)}{6^x(2^x+3^x)}=\frac{4^x-6^x+9^x}{6^x}=\left(\frac{2}{3}\right)^x-1+\left(\frac{3}{2}\right)^x=\frac{7}{6}$ Now let $u=\left(\frac{2}{3}\right)^x$ . Then $\frac{1}{u}=\left(\frac{3}{2}\right)^x$ , so that the last equation becomes $u-1+\frac{1}{u}=\frac{7}{6}$ Multiplying both sides by $6u$ will give us a quadratic equation in $u$ .Thus $6u^2-6u+6=7u$ $6u^2-13u+6=0$ $(2u-3)(3u-2)=0$ $u=\frac{3}{2}=\left(\frac{2}{3}\right)^{-1} \text{ or } u=\frac{2}{3}$ Now, since we have $u=\left(\frac{2}{3}\right)^x$ , we obtain $\frac{2}{3}=\left(\frac{2}{3}\right)^x$ or $x=1$ , and $\left(\frac{2}{3}\right)^{-1}=\left(\frac{2}{3}\right)^x$ or $x=-1$ . Let $a=1$ and $b=-1$ . Then $\sqrt{2a^2+2b^2}=\sqrt{2(1)^2+2(-1)^2}=\boxed{2}$ .

Well, if we try x = 1, it immediately works out. So, it must be true for x = -1 as well (why? because 8•27 = 12•18) , and since the problem says there's [only] two integer solutions, that's it.

8=2^3 and 27=3^3 and 12=3.2^2 and 18=2.3^2 now to minimize the equation; divide the equations by 6^x then taking y=(2/3)^x and solving it will provide answer 1, -1 and hence the answer of question is 2