Question Number 120120 by benjo_mathlover last updated on 29/Oct/20
$$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)\:=\:\mathrm{8}}\\{\mathrm{log}\:_{{b}} \left({x}\right)\:=\:\mathrm{3}\:}\\{\mathrm{log}\:_{{c}} \left({x}\right)\:=\:\mathrm{6}}\end{cases}\:\Rightarrow\:\mathrm{log}\:_{{abc}} \:\left({x}\right)=? \\ $$
Answered by bemath last updated on 29/Oct/20
$$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)=\mathrm{8}\Rightarrow\mathrm{log}\:_{{x}} \left({a}\right)=\frac{\mathrm{1}}{\mathrm{8}}}\\{\mathrm{log}\:_{{b}} \left({x}\right)=\mathrm{3}\Rightarrow\:\mathrm{log}\:_{{x}} \left({b}\right)=\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{log}\:_{{c}} \left({x}\right)=\mathrm{6}\Rightarrow\mathrm{log}\:_{{x}} \left({c}\right)=\frac{\mathrm{1}}{\mathrm{6}}}\end{cases} \\ $$$${Then}\:\mathrm{log}\:_{{abc}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{{x}} \left({abc}\right)}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{24}}{\mathrm{3}+\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{24}}{\mathrm{15}}\:=\:\frac{\mathrm{8}}{\mathrm{5}} \\ $$
Answered by $@y@m last updated on 29/Oct/20
$${x}={a}^{\mathrm{8}} ={b}^{\mathrm{3}} ={c}^{\mathrm{6}} \\ $$$${a}={x}^{\frac{\mathrm{1}}{\mathrm{8}}} ,\:\:{b}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} ,\:{c}={x}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${abc}={x}^{\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}} ={x}^{\frac{\mathrm{15}}{\mathrm{24}}} \\ $$$$\mathrm{log}_{{x}} \:\left({abc}\right)=\frac{\mathrm{15}}{\mathrm{24}} \\ $$$$\mathrm{log}\:_{{abc}} \:\left({x}\right)=\frac{\mathrm{24}}{\mathrm{15}}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$