The sum of all the solutions of the equation $$(8)^{2x} - 16 \cdot (8)^x + 48 = 0$$ is :

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

Let $$y=8^x$$---(i); 

Therefore, the equation can also be written as $$y^2-16y+48$$

 Factorize the equation: $$\left(y-12\right)\left(y-4\right)$$

Hence, replace values of y in the equation, (i) we get $$4=8^{x\ }and\ 12=8^x$$

Now take the log with base 8 on both sides. 

$$\therefore$$ $$x_1=\log_84$$ and $$x_2=\log_812$$

 Sum of all solutions $$\Rightarrow$$ $$x_1+x_2=\log_8\left(12\right)\left(4\right)$$

It can be also written as $$\log_88+\log_{_8}6$$ $$\Rightarrow$$ $$1+\log_{_8}6$$

Hence, the correct answer is option 1.