$${given}\:{that}\:{a},{b}\:{and}\:{c}\:{are}\:{positive}\:{numbers}\:{other}\:{than}\:\mathrm{1} \\$$$$,\:{show}\:{that}\:\:{log}_{{b}} {a}\:×\:{log}_{{c}} {b}\:×\:{log}_{{a}} {c}\:=\:\mathrm{1} \\$$$${hence},\:{evaluate}\:\:\:{log}_{\mathrm{10}} \mathrm{25}\:×\:{log}_{\mathrm{2}} \mathrm{10}\:×\:{log}_{\mathrm{5}} \mathrm{4} \\$$
Answered by $@ty@m123 last updated on 14/Sep/19 $${Let}\:\mathrm{log}_{{b}} \:{a}={x},\:\:{log}_{{c}} {b}={y},\:{log}_{{a}} {c}={z} \\$$$$\Rightarrow{b}^{{x}} ={a},\:{c}^{{y}} ={b},\:{a}^{{z}} ={c} \\$$$${Now}, \\$$$${a}^{{z}} ={c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{{Note}:\:{you}\:{can}\:{proceed}}\\{{with}\:{b}^{{x}} ={a}\:{or}\:{c}^{{y}} ={b}\:{also}.}\end{bmatrix} \\$$$$\Rightarrow\left({b}^{{x}} \right)^{{z}} ={c}\:\:\:\: \\$$$$\Rightarrow\left({c}^{{y}} \right)^{{xz}} ={c} \\$$$$\Rightarrow{c}^{{xyz}} ={c} \\$$$$\Rightarrow{xyz}=\mathrm{1} \\$$$$\Rightarrow{log}_{{b}} {a}\:×\:{log}_{{c}} {b}\:×\:{log}_{{a}} {c}\:=\:\mathrm{1} \\$$ Answered by$@ty@m123 last updated on 14/Sep/19
$${Hence} \\$$$${log}_{\mathrm{10}} \mathrm{25}\:×\:{log}_{\mathrm{2}} \mathrm{10}\:×\:{log}_{\mathrm{5}} \mathrm{4} \\$$$$=\mathrm{2log}_{\mathrm{10}} \:\mathrm{5}×{log}_{\mathrm{2}} \mathrm{10}\:×\mathrm{2}\:{log}_{\mathrm{5}} \mathrm{2} \\$$$$=\mathrm{4}\left(\mathrm{log}_{\mathrm{10}} \:\mathrm{5}×{log}_{\mathrm{2}} \mathrm{10}\:×{log}_{\mathrm{5}} \mathrm{2}\right) \\$$$$=\mathrm{4}×\mathrm{1}=\mathrm{4} \\$$
$${thanks}\:{so}\:{much} \\$$