l o g 1 10 8 + l o g 1 10 5 − l o g 1 10 2 {\displaystyle log_{\frac {1}{10}}{\sqrt {8}}+log_{\frac {1}{10}}{\sqrt {5}}-log_{\frac {1}{10}}2}
l o g 1 10 8 + l o g 1 10 5 − l o g 1 10 2 l o g 1 10 ( 8 + 5 − 2 ) l o g 1 10 ( 2 3 2 + 5 1 2 − 2 ) l o g 1 10 ( 2 2 2 ∗ 2 1 2 + 5 1 2 − 2 ) l o g 1 10 ( 2 1 2 + 5 1 2 ) a b = x ⇔ log a ( x ) = b 1 10 = ( 2 ∗ 5 ) 1 2 ( 10 ) − x = ( 2 ∗ 5 ) 1 2 ( 2 ∗ 5 ) − x = ( 2 ∗ 5 ) 1 2 x = − 1 2 {\displaystyle {\begin{aligned}log_{\frac {1}{10}}{\sqrt {8}}+log_{\frac {1}{10}}{\sqrt {5}}-log_{\frac {1}{10}}2\\log_{\frac {1}{10}}({\sqrt {8}}+{\sqrt {5}}-2)\\log_{\frac {1}{10}}(2^{\frac {3}{2}}+5^{\frac {1}{2}}-2)\\log_{\frac {1}{10}}(2^{\frac {2}{2}}*2^{\frac {1}{2}}+5^{\frac {1}{2}}-2)\\log_{\frac {1}{10}}(2^{\frac {1}{2}}+5^{\frac {1}{2}})\\{\displaystyle a^{b}=x\Leftrightarrow \log _{a}(x)=b}\\{\frac {1}{10}}=(2*5)^{\frac {1}{2}}\\(10)^{-x}=(2*5)^{\frac {1}{2}}\\(2*5)^{-x}=(2*5)^{\frac {1}{2}}\\x=-{\frac {1}{2}}\\\end{aligned}}}