Question:

Step 1: Simplify the right-hand side. $$\log_{\frac{1}{\sqrt{3}}}\left(\frac{1}{\sqrt{27}}\right) = \frac{\log(1/\sqrt{27})}{\log(1/\sqrt{3})} = \frac{-\frac{3}{2}\log 3}{-\frac{1}{2}\log 3} = 3$$ Step 2: Let $t = 2^x$. The equation becomes: $$\log_3(9t + 9) \cdot \log_3(t + 1) = 3$$ $$\log_3[9(t+1)] \cdot \log_3(t+1) = 3$$ $$(\log_3 9 + \log_3(t+1)) \cdot \log_3(t+1) = 3$$ $$(2 + \log_3(t+1)) \cdot \log_3(t+1) = 3$$ Step 3: Let $u = \log_3(t+1)$. Then: $$(2 + u) \cdot u = 3$$ $$u^2 + 2u - 3 = 0$$ $$(u+3)(u-1) = 0$$ $$u = 1 \text{ or } u = -3$$ Step 4: Solve for $x$. - If $u = 1$: $\log_3(2^x + 1) = 1 \Rightarrow 2^x + 1 = 3 \Rightarrow 2^x = 2 \Rightarrow x = 1$ - If $u = -3$: $\log_3(2^x + 1) = -3 \Rightarrow 2^x + 1 = 3^{-3} = \frac{1}{27} \Rightarrow 2^x = \frac{1}{27} - 1 < 0$, which is impossible. Step 5: Therefore $\alpha = 1$. Since $\alpha = 1$, it is less than $2$, $3$, and $4$.