Question Number 143965 by Khalmohmmad last updated on 20/Jun/21
Commented by Canebulok last updated on 20/Jun/21
$$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\frac{{log}\left({x}\right)}{{log}\left(\mathrm{3}\right)}\:+\:\frac{{log}\left(\mathrm{5}\right)}{{log}\left({x}\right)}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{log}\left({x}\right)^{\mathrm{2}} \:+\:{log}\left(\mathrm{5}\right){log}\left(\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$$${Let}:\:\:{k}\:=\:{log}\left({x}\right) \\ $$$$\Rightarrow\:{k}^{\mathrm{2}} \:+\:{log}\left(\mathrm{5}\right){log}\left(\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$${By}\:{using}\:{the}\:{quadratic}\:{formula}: \\ $$$$\Rightarrow\:{k}\:=\:\frac{\pm\sqrt{−\mathrm{4}\left({log}\left(\mathrm{3}\right){log}\left(\mathrm{5}\right)\right)}}{\mathrm{2}} \\ $$$$\Rightarrow\:{k}_{\mathrm{1}} \:=\:+\:{i}\sqrt{{log}\left(\mathrm{3}\right){log}\left(\mathrm{5}\right)} \\ $$$$\Rightarrow\:{k}_{\mathrm{2}} \:=\:−{i}\sqrt{{log}\left(\mathrm{3}\right){log}\left(\mathrm{5}\right)} \\ $$$$\: \\ $$$${Thus}; \\ $$$$\Rightarrow\:{log}\left({x}\right)\:=\:\pm{i}\sqrt{{log}\left(\mathrm{3}\right){log}\left(\mathrm{5}\right)} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{10}^{\pm{i}\sqrt{{log}\left(\mathrm{3}\right){log}\left(\mathrm{5}\right)}} \\ $$$$\: \\ $$$$\sim\:{Kevin} \\ $$
Answered by mr W last updated on 20/Jun/21
$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{3}}+\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:{x}}=\mathrm{0} \\ $$$$\mathrm{ln}\:{x}=\pm{i}\sqrt{\mathrm{ln}\:\mathrm{3}\:\mathrm{ln}\:\mathrm{5}} \\ $$$${x}={e}^{\pm{i}\sqrt{\mathrm{ln}\:\mathrm{3}\:\mathrm{ln}\:\mathrm{5}}} =\mathrm{cos}\:\sqrt{\mathrm{ln}\:\mathrm{3}\:\mathrm{ln}\:\mathrm{5}}\pm{i}\:\mathrm{sin}\:\sqrt{\mathrm{ln}\:\mathrm{3}\:\mathrm{ln}\:\mathrm{5}} \\ $$