7.305 Simplify the expression $A=\left(\left(\log _{b}^{4} a+\log _{a}^{4} b+2\right)^{\frac{1}{2}}-2\right)^{\frac{1}{2}}$ for $1<\boldsymbol{a}<\boldsymbol{b}$.
Solution.

$$
\begin{aligned}
& A=\left(\left(\log _{b}^{4} a+2 \log _{a}^{2} b \cdot \log _{b}^{2} a+\log _{a}^{4} b\right)^{\frac{1}{2}}-2\right)^{\frac{1}{2}}=\left(\log _{b}^{2} a+\log _{a}^{2} b-2\right)^{\frac{1}{2}}= \\
& =\left(\log _{b}^{2} a-2 \log _{a} b \cdot \log _{b} a+\log _{a}^{2} b\right)^{\frac{1}{2}}=\left|\log _{b} a-\log _{a} b\right| \\
& \log _{a} b>1(b>a>1) \Rightarrow \log _{a} b>\frac{1}{\log _{a} b} \Rightarrow \log _{a} b>\log _{b} a
\end{aligned}
$$

Therefore, $A=\log _{a} b-\log _{b} a$.

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

## In the solutions to problems 7.306 - 7.307, the method described in detail in section XVII of the reference materials is used.