7.014. $\left(\log _{a} b+\log _{b} a+2\right)\left(\log _{a} b-\log _{a b} b\right) \log _{b} a-1$.
Solution.

$$
\begin{aligned}
& \left(\log _{a} b+\log _{b} a+2\right)\left(\log _{a} b-\log _{a b} b\right) \log _{b} a-1=\left(\log _{a} b+\frac{1}{\log _{a} b}+2\right) \times \\
& \times\left(\log _{a} b-\frac{\log _{a} b}{\log _{a} a b}\right) \frac{1}{\log _{a} b}-1=\frac{\log _{a}^{2} b+2 \log _{a} b+1}{\log _{a} b} \times \\
& \times\left(\log _{a} b-\frac{\log _{a} b}{\log _{a} a+\log _{a} b}\right) \cdot \frac{1}{\log _{a} b}-1= \\
& =\frac{\left(\log _{a} b+1\right)^{2}}{\log _{a} b} \cdot\left(\log _{a} b-\frac{\log _{a} b}{1+\log _{a} b}\right) \frac{1}{\log _{a} b}-1= \\
& =\frac{\left(\log _{a} b+1\right)^{2}}{\log _{a} b} \log _{a} b\left(1-\frac{1}{1+\log _{a} b}\right) \frac{1}{\log _{a} b}-1=\frac{\left(\log _{a} b+1\right)^{2}\left(1+\log _{a} b-1\right)}{\left(1+\log _{a} b\right) \log _{a} b}- \\
& -1=\log _{a} b+1-1=\log _{a} b
\end{aligned}
$$

Answer: $\log _{a} b$.