1. **State the problem:** We are given $\log_a 2 = t$ and need to express $\log_{\sqrt{2a}} 8$ in terms of $t$. 2. **Recall the change of base formula:** For any positive numbers $x, b, c$ with $b \neq 1$ and $c \neq 1$, $$\log_b x = \frac{\log_c x}{\log_c b}$$ 3. **Rewrite the base and argument:** - The base is $\sqrt{2a} = (2a)^{1/2}$. - The argument is $8 = 2^3$. 4. **Apply the change of base formula using base $a$:** $$\log_{\sqrt{2a}} 8 = \frac{\log_a 8}{\log_a \sqrt{2a}}$$ 5. **Express numerator and denominator in terms of $t$:** - Numerator: $$\log_a 8 = \log_a 2^3 = 3 \log_a 2 = 3t$$ - Denominator: $$\log_a \sqrt{2a} = \log_a (2a)^{1/2} = \frac{1}{2} \log_a (2a) = \frac{1}{2} (\log_a 2 + \log_a a)$$ Since $\log_a a = 1$, and $\log_a 2 = t$, we get $$\frac{1}{2} (t + 1)$$ 6. **Combine numerator and denominator:** $$\log_{\sqrt{2a}} 8 = \frac{3t}{\frac{1}{2} (t + 1)} = \frac{3t}{\frac{t+1}{2}} = 3t \times \frac{2}{t+1} = \frac{6t}{t+1}$$ **Final answer:** $$\boxed{\log_{\sqrt{2a}} 8 = \frac{6t}{t+1}}$$