l o g 1 8 ( 32 2 ) {\displaystyle log_{\frac {1}{8}}({\frac {32}{\sqrt {2}}})}
על פי כלל הלוגרתם a b = x ⇔ log a ( x ) = b {\displaystyle a^{b}=x\Leftrightarrow \log _{a}(x)=b}
l o g 1 8 ( 32 2 ) = x l o g 1 8 ( 2 5 2 1 2 ) = x l o g 1 8 ( 2 5 − 1 2 ) = x l o g 1 8 ( 2 4.5 ) = x 2 4.5 = 1 8 x 2 4.5 = 8 − x 2 4.5 = 2 − 3 x 4.5 = − 3 x x = − 1.5 {\displaystyle {\begin{aligned}log_{\frac {1}{8}}({\frac {32}{\sqrt {2}}})=x\\log_{\frac {1}{8}}({\frac {2^{5}}{2{\frac {1}{2}}}})=x\\log_{\frac {1}{8}}(2^{5-{\frac {1}{2}}})=x\\log_{\frac {1}{8}}(2^{4.5})=x\\2^{4.5}={\frac {1}{8}}^{x}\\2^{4.5}=8^{-x}\\2^{4.5}=2^{-3x}\\4.5=-3x\\x=-1.5\\\end{aligned}}}
