7.155. $\left(6\left(\log _{b} a \cdot \log _{a^{2}} b+1\right)+\log _{a} b^{-6}+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b$ for $a>1$.
Solution.

$$
\begin{aligned}
& \left(6\left(\log _{b} a \cdot \log _{a^{2}} b+1\right)+\log _{a} b^{-6}+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b= \\
& =\left(6\left(\frac{1}{2}+1\right)-6 \log _{a} b+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b=\sqrt{9-6 \log _{a} b+\log _{a}^{2} b}- \\
& -\log _{a} b=\sqrt{\left(3-\log _{a} b\right)^{2}}-\log _{a} b=\left|3-\log _{a} b\right|-\log _{a} b
\end{aligned}
$$

By expanding the absolute value, we get two cases:

1) $\left|3-\log _{a} b\right|-\log _{a} b=\left\{\begin{array}{l}3-\log _{a} b \leq 0, \\ -3+\log _{a} b-\log _{a} b=-3 ;\end{array}\right.$

$\left\{\begin{array}{l}b \geq a^{3} \\ \left|3-\log _{a} b\right|-\log _{a} b=-3\end{array}\right.$
2) $\left|3-\log _{a} b\right|-\log _{a} b=\left\{\begin{array}{l}3-\log _{a} b>0, \\ 3-\log _{a} b-\log _{a} b=3-2 \log _{a} b ;\end{array}\right.$

$\left\{\begin{array}{l}0<b<a^{3}, b \neq 1, \\ \left|3-\log _{a} b\right|-\log _{a} b=3-2 \log _{a} b .\end{array}\right.$

Answer: -3 if $b \geq a^{3}$, and $3-2 \log _{a} b$ if $0<b<a^{3}, b \neq 1$.