Question Number 79978 by mr W last updated on 29/Jan/20
$${Given}\:{for}\:{x},{y},{z}>\mathrm{0}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \\ $$$${Arrange}\:\mathrm{2}{x},\:\mathrm{3}{y},\:\mathrm{5}{z}\:{in}\:{increasing}\:{order}. \\ $$
Answered by mind is power last updated on 29/Jan/20
$${xln}\left(\mathrm{2}\right)={yln}\left(\mathrm{3}\right)={zln}\left(\mathrm{5}\right) \\ $$$${y}=\frac{{xln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$${z}=\frac{{xln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)} \\ $$$$\mathrm{5}{z}=\frac{\mathrm{5}}{{ln}\left(\mathrm{5}\right)}.{ln}\left(\mathrm{2}\right){x} \\ $$$$\mathrm{3}{y}=\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}{x} \\ $$$${ln}\left(\mathrm{8}\right)<{ln}\left(\mathrm{9}\right)\Rightarrow \\ $$$$\mathrm{3}{ln}\left(\mathrm{2}\right)<\mathrm{2}{ln}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}<\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}{y}=\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}{x}<\mathrm{2}{x} \\ $$$$\frac{\mathrm{5}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)}>\mathrm{2}\Leftrightarrow{ln}\left(\mathrm{32}\right)>{ln}\left(\mathrm{25}\right)\:{True}\Rightarrow \\ $$$$\mathrm{5}{z}=\frac{\mathrm{5}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)}{x}>\mathrm{2}{x}\Rightarrow\mathrm{5}{z}>\mathrm{2}{x}>\mathrm{3}{y} \\ $$
Answered by key of knowledge last updated on 29/Jan/20
$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \Rightarrow\mathrm{x}.\mathrm{log}_{\mathrm{3}} \mathrm{2}=\mathrm{y}=\mathrm{z}.\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\mathrm{2x}=\mathrm{2}\left(\frac{\mathrm{y}}{\mathrm{1og}_{\mathrm{3}} \mathrm{2}}\right)=\mathrm{y}.\mathrm{2log}_{\mathrm{2}} \mathrm{3}\approx\mathrm{3}.\mathrm{16y} \\ $$$$\mathrm{5z}=…=\mathrm{z}×\mathrm{5log}_{\mathrm{5}} \mathrm{3}\approx\mathrm{3}.\mathrm{41y} \\ $$$$\Rightarrow\mathrm{3y}<\mathrm{2x}<\mathrm{5z} \\ $$
Answered by mr W last updated on 30/Jan/20
$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} ={t}>\mathrm{0} \\ $$$${x}\:\mathrm{ln}\:\mathrm{2}={y}\:\mathrm{ln}\:\mathrm{3}={z}\:\mathrm{ln}\:\mathrm{5}=\mathrm{ln}\:{t}={s} \\ $$$$\Rightarrow{x}=\frac{{s}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{x}=\frac{{s}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}}=\frac{{s}}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{ln}\:\sqrt{\mathrm{2}}}{{s}} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{y}}=\frac{\mathrm{ln}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}}{{s}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{5}{z}}=\frac{\mathrm{ln}\:\sqrt[{\mathrm{5}}]{\mathrm{5}}}{{s}} \\ $$$${now}\:{we}\:{only}\:{need}\:{to}\:{compare}\:{the} \\ $$$${numbers}\:\sqrt{\mathrm{2}},\:\sqrt[{\mathrm{3}}]{\mathrm{3}},\:\sqrt[{\mathrm{5}}]{\mathrm{5}}. \\ $$$$\mathrm{3}^{\mathrm{2}} =\mathrm{9}>\mathrm{8}=\mathrm{2}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}>\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{5}} =\mathrm{32}>\mathrm{25}=\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}>\left(\sqrt[{\mathrm{5}}]{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}>\sqrt[{\mathrm{5}}]{\mathrm{5}} \\ $$$$ \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt{\mathrm{2}}>\sqrt[{\mathrm{5}}]{\mathrm{5}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{y}}>\frac{\mathrm{1}}{\mathrm{2}{x}}>\frac{\mathrm{1}}{\mathrm{5}{z}} \\ $$$$\Rightarrow\mathrm{3}{y}<\mathrm{2}{x}<\mathrm{5}{z} \\ $$