Question Number 191839 by MATHEMATICSAM last updated on 01/May/23
$$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$
Answered by deleteduser1 last updated on 01/May/23
$${a}={blog}_{\mathrm{2}} \left(\mathrm{3}\right)\Rightarrow\frac{{a}}{{b}}={log}_{\mathrm{2}} \left(\mathrm{3}\right)\Rightarrow\frac{{a}+{b}}{{b}}={log}_{\mathrm{2}} \left(\mathrm{6}\right) \\ $$$${a}={clog}_{\mathrm{2}} \left(\mathrm{36}\right)\Rightarrow\frac{{a}}{{c}}={log}_{\mathrm{2}} \left(\mathrm{36}\right)\Rightarrow\frac{{a}}{\mathrm{2}{c}}={log}_{\mathrm{2}} \left(\mathrm{6}\right) \\ $$$$\Rightarrow\frac{{a}}{\mathrm{2}{c}}=\frac{{a}+{b}}{{b}}\Rightarrow{ab}=\mathrm{2}{c}\left({a}+{b}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$