We use the following three rules for logs :

1. $n\log_a(m)=\log_a(m^n)$

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

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

Using rule 1 we get
$$2\log_{\var{a}}(\simplify{x+{b}})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}((\simplify{x+{b}})^2)- \log_{\var{a}}(\simplify{(x+{c})})$$
Using rule 2 gives
$$\log_{\var{a}}(\simplify{(x+{b})^2})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}\left(\simplify{(x+{b})^2/(x+{c})}\right)$$
So the equation to solve becomes:
$$\log_{\var{a}}\left(\simplify{(x+{b})^2/(x+{c})}\right)=\var{d}$$
and using rule 3 this gives:
$$\begin{eqnarray} \simplify{(x+{b})^2/(x+{c})}&=&{\var{a}}^{\var{d}}\Rightarrow\\ \simplify{(x+{b})^2}&=&{\var{a}}^{\var{d}}(\simplify{x+{c}})=\simplify{{a^d}(x+{c})}\Rightarrow\\ \simplify{x^2+{2*b-a^(d)}x+{b^2-a^(d)*c}}&=&0 \end{eqnarray}$$
Solving this quadratic we get two solutions:

$x=\var{sol1}$ and $x=\var{sol2}$

We should check that these solutions gives positive values for $\simplify{x+{b}}$ and $\simplify{x+{c}}$ as otherwise the logs are not defined.

The value $x=\var{sol1}$ gives: 

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{b}})$ we get $\log_{\var{a}}(\simplify{{2*a^d}})$ so OK.

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{c}})$ we get $\log_{\var{a}}(\simplify{{4*a^d}})$ so OK.

Hence $x=\var{sol1}$ is a solution to our original equation.

The value $x=\var{sol2}$ gives:

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{b}})$ we get $\log_{\var{a}}(\simplify{{-a^d}})$ so NOT OK.

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

Hence $x=\var{sol2}$ is NOT a solution to our original equation as $\log_{\var{a}}(\simplify{x+{b}})$ is not defined for this value of $x$.

So there is only one solution $x=\var{sol1}$.