Remember the basic rules of Logarithm

$\begin{aligned}\begin{aligned}3\log_8x=&\ \log_{4}(x+6)\\3\log_{2^3}x=&\ \log_{2^2}(x+6)\\3\cdot\frac13\log_2x=&\ \frac12\log_{2}(x+6)\\\log_2x=&\ \log_{2}(x+6)^{\frac12}\\x=&\ (x+6)^{\frac12}\\x^2=&\ x+6\\x^2-x-6=&\ 0\\(x-3)(x+2)=&\ 0\\x=&\ 3,-2\end{aligned}\end{aligned}$

Since the logarithm functions $\log_8x$ and $\log_4{(x+6)}$ are defined over positive numbers, the values of $x+6$ and $x$ are positive. Thus, $-2$ is can not be the value of $x$ implying that the value of $3\log_8x=\log_{4}(x+6)$ satisfying is $x=\boxed{3}$

the blue curve is $y=\log _4\left(x+6\right)$ and red curve is $y=3\log _8\left(x\right)$ . These intersect at $(3,1.585)$

3 is the only solution. This logarithmic equation can be converted to a quadratic equation by writing it is log to the base2 and then eliminating the logarithm.

We get x^2 - x - 6 = 0

Too early to conclude that the sum of the roots is 1... The two roots are -2 and 3. when we plug in -2 in the given equation, it becomes invalid. Hence 3 is the only solution and the sum of roots is 3.