Question Number 110183 by bemath last updated on 27/Aug/20
$$\:\:\:\sqrt{\bigstar}\frac{{be}}{{math}}\sqrt{\bigstar} \\ $$$$\:\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right)+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{log}\:_{\mathrm{4}} \left({x}\right)=\mathrm{1} \\ $$$$\:\:\:{x}=? \\ $$
Answered by Olaf last updated on 27/Aug/20
$$\mathrm{log}_{\mathrm{2}} {x}+\mathrm{log}_{\mathrm{3}} {x}+\mathrm{log}_{\mathrm{4}} {x}\:=\:\mathrm{1} \\ $$$$\frac{\mathrm{ln}{x}}{\mathrm{ln2}}+\frac{\mathrm{ln}{x}}{\mathrm{ln3}}+\frac{\mathrm{ln}{x}}{\mathrm{ln4}}\:=\:\mathrm{1} \\ $$$$\mathrm{ln}{x}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{ln2}}+\frac{\mathrm{1}}{\mathrm{ln3}}+\frac{\mathrm{1}}{\mathrm{ln4}}} \\ $$$${x}\:=\:{e}^{\mathrm{1}/\left(\frac{\mathrm{1}}{\mathrm{ln2}}+\frac{\mathrm{1}}{\mathrm{ln3}}+\frac{\mathrm{1}}{\mathrm{ln4}}\right)} \:=\:{e}^{\mathrm{1}/\left(\frac{\mathrm{3}}{\mathrm{2ln2}}+\frac{\mathrm{1}}{\mathrm{ln3}}\right)} \\ $$$$ \\ $$
Answered by Aziztisffola last updated on 27/Aug/20
$$\:\mathrm{ln}\left(\mathrm{x}\right)\left(\frac{\mathrm{1}}{\mathrm{ln2}}+\frac{\mathrm{1}}{\mathrm{ln3}}+\frac{\mathrm{1}}{\mathrm{ln4}}\right)=\mathrm{1} \\ $$$$\:\mathrm{ln}\left(\mathrm{x}\right)\left(\frac{\mathrm{ln3ln4}+\mathrm{ln2ln4}+\mathrm{ln2ln3}}{\mathrm{ln2ln3ln4}}\right)=\mathrm{1} \\ $$$$\:\mathrm{ln}\left(\mathrm{x}\right)=\frac{\mathrm{ln2ln3ln4}}{\mathrm{ln3ln4}+\mathrm{ln2ln4}+\mathrm{ln2ln3}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}\:\:\:=\mathrm{e}^{\frac{\mathrm{ln2ln3ln4}}{\mathrm{ln3ln4}+\mathrm{ln2ln4}+\mathrm{ln2ln3}}} \\ $$
Answered by john santu last updated on 28/Aug/20
![log _2 (x)+((log _2 (x))/(log _2 (3)))+((log _2 (x))/(log _2 (4)))=1 log _2 (x)[1+(1/(log _2 (3)))+(1/2)]=1 log _2 (x) [(3/2)+(1/(log _2 (3)))]=1 log _2 (x)(((log _2 (27)+log _2 (4))/(log _2 (9))))=1 log _2 (x)(((log _2 (108))/(log _2 (9))))=1 log _2 (x)=((log _2 (9))/(log _2 (108))) log _2 (x)=log _(108) (9) ⇒ x = 2^(log _(108) (9)) .△](Q110251.png)
$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)+\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)}+\frac{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{4}\right)}=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\right]=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\:\left[\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)}\right]=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\left(\frac{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{27}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{4}\right)}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{9}\right)}\right)=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\left(\frac{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{108}\right)}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{9}\right)}\right)=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{9}\right)}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{108}\right)} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{log}\:_{\mathrm{108}} \left(\mathrm{9}\right) \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{2}^{\mathrm{log}\:_{\mathrm{108}} \left(\mathrm{9}\right)} \:.\bigtriangleup \\ $$