Exponent vs Logarithm

CaseI : $3^{x}\geq1$

$3^{x}+1-3^{x}+1=2log_{5}|6-x|$

$1=log_{5}|6-x|$

This gives $x=1,11$

CaseII : $3^{x}\leq1$

$3^{x}+1+3^{x}-1=2log_{5}|6-x|$

$3^{x}=log_{5}|6-x|$

Here no solution does not exist since $log_{5}|6-x|$ is always greater than one because $log_{5}6<log_{5}|6-x|(since..x<0)$ .

Hence sum of solutions of $x$ is $\boxed{12}$ .

2 cases are needed to be considered $3^x>1 \Rightarrow x>0$ and $3^x<1 \Rightarrow x<0$

I $x>0 \Rightarrow 2=2 \log_5 |x-6| \Rightarrow x=6 \pm 5 \Rightarrow x =1,11$ II $x<0 \Rightarrow 2*3^x=2 \log_5 |6-x|$ $\Rightarrow$ x=6-5^3^x>1 $\Rightarrow x∈∅$ Combining x=1,11......Sum=12

We can break the equation in cases...subsequently 3 cases would be made one for x<=0 then for x<6 and x>6 and as we find x=1 satisfies the equation then by symmetry along x=6 line x=11 would also satisfy..hence 1+11=12