Question Number 22928 by selestian last updated on 24/Oct/17
Answered by $@ty@m last updated on 24/Oct/17
$${Let}\:{x}^{\mathrm{18}} ={y}^{\mathrm{21}} ={z}^{\mathrm{28}} ={k} \\ $$$$\Rightarrow{x}={k}^{\frac{\mathrm{1}}{\mathrm{18}}} ,\:{y}={k}^{\frac{\mathrm{1}}{\mathrm{21}}} ,{z}={k}^{\frac{\mathrm{1}}{\mathrm{28}}} \\ $$$$\therefore\:\mathrm{3log}\:_{{y}} {x}=\mathrm{3}\frac{\mathrm{log}\:{x}}{\mathrm{log}\:{y}}=\mathrm{3}\frac{\mathrm{log}\:{k}^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{log}\:{k}^{\frac{\mathrm{1}}{\mathrm{21}}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\frac{\frac{\mathrm{1}}{\mathrm{18}}}{\frac{\mathrm{1}}{\mathrm{21}}}=\mathrm{3}×\frac{\mathrm{21}}{\mathrm{18}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${Similarly} \\ $$$$\mathrm{3log}\:_{{z}} {y}=\mathrm{3}×\frac{\mathrm{28}}{\mathrm{21}}=\mathrm{4} \\ $$$${and}\:\mathrm{7log}\:_{{x}} {z}=\mathrm{7}×\frac{\mathrm{18}}{\mathrm{28}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{3},\frac{\mathrm{7}}{\mathrm{2}},\mathrm{4},\frac{\mathrm{9}}{\mathrm{2}}\:{are}\:{in}\:{AP} \\ $$