1. Substitute to reduce the equation to a quadratic

Given: $8^{2x}-16\cdot 8^x+48=0$

Let $t=8^x$ Then $t>0$, and the equation becomes $t^2-16t+48=0$

2. Solve the quadratic

Factorizing: $t^2-16t+48=(t-12)(t-4)=0$

So, $t=12 \quad \text{or} \quad t=4$

Since $t=8^x$, we get: $8^x=12 \quad \text{or} \quad 8^x=4$

Thus, $x=\log_8(12) \quad \text{or} \quad x=\log_8(4)$

3. Find the sum of the solutions

Sum: $\log_8(12)+\log_8(4)=\log_8(12\cdot 4)=\log_8(48)$

Now, $48=8\cdot 6$ so $\log_8(48)=\log_8(8\cdot 6)=\log_8 8+\log_8 6=1+\log_8 6$

Therefore, the sum of all solutions is $1+\log_8(6)$

4. Check options

• A: $1+\log_8(6)$ ✅
• B: $1+\log_6(8)$ ❌
• C: $\log_8(6)$ ❌
• D: $\log_8(4)$ ❌

Hence, the correct option is A.