Question Number 172354 by Mikenice last updated on 25/Jun/22
Answered by mahdipoor last updated on 25/Jun/22
$${log}_{{a}} {b}={m}\:\:\:\:{log}_{{a}} {c}={n} \\ $$$$\Rightarrow\begin{cases}{{log}_{{n}^{\mathrm{2}} } \left(\mathrm{1}/{m}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{log}_{{n}} {m}=\frac{−\mathrm{3}}{\mathrm{2}}}\\{\frac{{log}_{{a}} {b}}{{log}_{{a}} {c}}−{n}=\frac{{m}}{{n}}−{n}=\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{n}^{\mathrm{3}} ={m}}\\{{m}=\mathrm{2}{n}+{n}^{\mathrm{2}} }\end{cases}\Rightarrow\begin{cases}{{n}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{m}=\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow{A}=\mathrm{2018}{m}+\mathrm{2019}{n}=\mathrm{6054}.\mathrm{5}+\mathrm{4036}.\mathrm{5}\sqrt{\mathrm{5}} \\ $$