1.Ta có:$\log_{25}80=\dfrac{1}{2}\log_580=\dfrac{1}{2}(1+4\log_52)$$b=\log_{\sqrt{27}}40=2\log_{27}40=2(\log_{27}8+\log_{27}5)=2(\log_32+\dfrac{1}{3}\log_35)$$a=\log_950=\dfrac{1}{2}\log_350=\dfrac{1}{2}(\log_32+2\log_25)$$\Rightarrow \left\{\begin{array}{l}\log_32=\dfrac{3b-2a}{5}\\\log_35=\dfrac{12a-3b}{10}\end{array}\right.$$\Rightarrow \log_52=\dfrac{\log_32}{\log_35}=\dfrac{6b-4a}{12a-3b}$$\Rightarrow \log_{25}80=\dfrac{1}{2}+\dfrac{12b-8a}{12a-3b}$
2.Ta có:$\log_{25}80=\dfrac{1}{2}\log_580=\dfrac{1}{2}(1+4\log_52)$$b=\log_{\sqrt{27}}40=2\log_{27}40=2(\log_{27}8+\log_{27}5)=2(\log_32+\dfrac{1}{3}\log_35)$$a=\log_950=\dfrac{1}{2}\log_350=\dfrac{1}{2}(\log_32+2\log_25)$$\Rightarrow \left\{\begin{array}{l}\log_32=\dfrac{3b-2a}{5}\\\log_35=\dfrac{12a-3b}{10}\end{array}\right.$$\Rightarrow \log_52=\dfrac{\log_32}{\log_35}=\dfrac{6b-4a}{12a-3b}$$\Rightarrow \log_{25}80=\dfrac{1}{2}+\dfrac{12b-8a}{12a-3b}$