C-04-Q235mediumsingle_mcq5logab+5logbc+5logca=?5 \log\dfrac{a}{b} + 5\log\dfrac{b}{c} + 5\log\dfrac{c}{a} = ?5logba+5logcb+5logac=?a000b111c555dlogabc\log abclogabcব্যাখ্যাThe arguments multiply to ab⋅bc⋅ca=1\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a} = 1ba⋅cb⋅ac=1, so the sum of logs is 5log1=05\log 1 = 05log1=0.