We use the following two rules for logs :

1. $\log_a(b)-\log_a(c)=\log_a(b/c)$

2. $\log_a(p)=r \Rightarrow p=a^r$

Using rule 1 we get
$$\log_{\var{a}}(x+\var{b})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}\left(\simplify{(x+{b})/(x+{c})}\right)$$
So the equation to solve becomes:
$$\log_{\var{a}}\left(\simplify{(x+{b})/(x+{c})}\right)=\var{d}$$
and using rule 2 this gives:
$$\begin{eqnarray} \simplify{(x+{b})/(x+{c})}&=&{\var{a}}^{\var{d}}\Rightarrow\\ x+\var{b}&=&{\var{a}}^{\var{d}}(x+\var{c})=\simplify{{a^d}}(x+\var{c})\Rightarrow\\ \simplify{{a^d-1}x}&=&\simplify[std]{{b}-{c}*{a^d}={b-c*a^d}}\Rightarrow\\ x&=&\simplify{{b-c*a^d}/{a^d-1}} \end{eqnarray}$$
We should check that this solution gives positive values for $x+\var{b}$ and $\simplify{x+{c}}$ as otherwise the logs are not defined.

Substituting this value for $x$ into $\log_{\var{a}}(x+\var{b})$ we get $\log_{\var{a}}(\simplify{({b-c }{a^d})/{a^d-1}})$ so OK.

For $\log_{\var{a}}(\simplify{x+{c}})$ we get on substituting for $x$, $\log_{\var{a}}(\simplify{({b-c })/{a^d-1}})$ so OK.

Hence the value we found for $x$ is a solution to the original equation.