Logarithmic functions?I am stuck on this question which as follows: log ⁡ ( x )...

The first approach is correct. The second is not: $\log (x-3)\ne \log x-\log 3$
The next step is to check for extraneous solutions, by verifying which solutions ( $x=0$ and $x=13$) satisfy the original equation.

I am not sure if $log(z)$ means $lo{g}_{10}(z)$ or $lo{g}_e(z)$ i.e $ln(z)$ so I will go with convention and assume $log(z)⟹lo{g}_{10}(z)$
Here goes:
so $x\in \{0,13\}$
But for $x=0$ the original equation becomes: $log(0)+log(-3)=log(0)$ his is a problem because $log(z)$ is defined only for $0<z$ if z is real.
Logarithms of complex numbers quickly get very messy. So confining the arguments of the log functions to positive reals leaves a single answer: $x=13$
I hope this was not too much detail.