Question Number 208412 by efronzo1 last updated on 15/Jun/24
Answered by A5T last updated on 15/Jun/24
$$\mathrm{8}=\mathrm{24}^{\frac{\mathrm{1}}{{a}}} ,\mathrm{27}=\mathrm{24}^{\frac{\mathrm{1}}{{b}}} ,\mathrm{64}=\mathrm{24}^{\frac{\mathrm{1}}{{c}}} \\ $$$$\Rightarrow\mathrm{24}^{\mathrm{3}} =\mathrm{8}×\mathrm{27}×\mathrm{64}=\mathrm{24}^{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{ab}+{bc}+{ca}}{{abc}}\right)} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}}{{abc}}=\mathrm{3}\Rightarrow\frac{\mathrm{2022}{abc}}{{ab}+{bc}+{ca}}=\mathrm{2022}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{674} \\ $$
Answered by MM42 last updated on 15/Jun/24
$${a}=\frac{\mathrm{1}}{\mathrm{3}{log}_{\mathrm{2}} \mathrm{24}}\:\&\:{b}=\frac{\mathrm{1}}{\mathrm{3}{log}_{\mathrm{3}} \mathrm{24}}\:\:\&\:\:{c}=\frac{\mathrm{1}}{\mathrm{3}{log}_{\mathrm{4}} \mathrm{24}} \\ $$$${A}=\frac{\mathrm{2022}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}=\frac{\mathrm{2022}}{\mathrm{3}\left({log}_{\mathrm{2}} \mathrm{24}+{log}_{\mathrm{3}} \mathrm{24}+{log}_{\mathrm{4}} \mathrm{24}\right)} \\ $$$$=\frac{\mathrm{2022}}{\mathrm{3}}=\:\mathrm{674}\:\checkmark \\ $$
Commented by efronzo1 last updated on 15/Jun/24
$$\mathrm{674} \\ $$
Commented by MM42 last updated on 15/Jun/24
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$