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Given-for-x-y-z-gt-0-2-x-3-y-5-z-Arrange-2x-3y-5z-in-increasing-order-




Question Number 79978 by mr W last updated on 29/Jan/20
Given for x,y,z>0:  2^x =3^y =5^z   Arrange 2x, 3y, 5z in increasing order.
$${Given}\:{for}\:{x},{y},{z}>\mathrm{0}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \\ $$$${Arrange}\:\mathrm{2}{x},\:\mathrm{3}{y},\:\mathrm{5}{z}\:{in}\:{increasing}\:{order}. \\ $$
Answered by mind is power last updated on 29/Jan/20
xln(2)=yln(3)=zln(5)  y=((xln(2))/(ln(3)))  z=((xln(2))/(ln(5)))  5z=(5/(ln(5))).ln(2)x  3y=((3ln(2))/(ln(3)))x  ln(8)<ln(9)⇒  3ln(2)<2ln(3)  ⇒((3ln(2))/(ln(3)))<2  ⇒3y=((3ln(2))/(ln(3)))x<2x  ((5ln(2))/(ln(5)))>2⇔ln(32)>ln(25) True⇒  5z=((5ln(2))/(ln(5)))x>2x⇒5z>2x>3y
$${xln}\left(\mathrm{2}\right)={yln}\left(\mathrm{3}\right)={zln}\left(\mathrm{5}\right) \\ $$$${y}=\frac{{xln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$${z}=\frac{{xln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)} \\ $$$$\mathrm{5}{z}=\frac{\mathrm{5}}{{ln}\left(\mathrm{5}\right)}.{ln}\left(\mathrm{2}\right){x} \\ $$$$\mathrm{3}{y}=\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}{x} \\ $$$${ln}\left(\mathrm{8}\right)<{ln}\left(\mathrm{9}\right)\Rightarrow \\ $$$$\mathrm{3}{ln}\left(\mathrm{2}\right)<\mathrm{2}{ln}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}<\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}{y}=\frac{\mathrm{3}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)}{x}<\mathrm{2}{x} \\ $$$$\frac{\mathrm{5}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)}>\mathrm{2}\Leftrightarrow{ln}\left(\mathrm{32}\right)>{ln}\left(\mathrm{25}\right)\:{True}\Rightarrow \\ $$$$\mathrm{5}{z}=\frac{\mathrm{5}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{5}\right)}{x}>\mathrm{2}{x}\Rightarrow\mathrm{5}{z}>\mathrm{2}{x}>\mathrm{3}{y} \\ $$
Answered by key of knowledge last updated on 29/Jan/20
2^x =3^y =5^z ⇒x.log_3 2=y=z.log_3 5  2x=2((y/(1og_3 2)))=y.2log_2 3≈3.16y  5z=...=z×5log_5 3≈3.41y  ⇒3y<2x<5z
$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \Rightarrow\mathrm{x}.\mathrm{log}_{\mathrm{3}} \mathrm{2}=\mathrm{y}=\mathrm{z}.\mathrm{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\mathrm{2x}=\mathrm{2}\left(\frac{\mathrm{y}}{\mathrm{1og}_{\mathrm{3}} \mathrm{2}}\right)=\mathrm{y}.\mathrm{2log}_{\mathrm{2}} \mathrm{3}\approx\mathrm{3}.\mathrm{16y} \\ $$$$\mathrm{5z}=…=\mathrm{z}×\mathrm{5log}_{\mathrm{5}} \mathrm{3}\approx\mathrm{3}.\mathrm{41y} \\ $$$$\Rightarrow\mathrm{3y}<\mathrm{2x}<\mathrm{5z} \\ $$
Answered by mr W last updated on 30/Jan/20
2^x =3^y =5^z =t>0  x ln 2=y ln 3=z ln 5=ln t=s  ⇒x=(s/(ln 2))  ⇒2x=(s/((1/2)ln 2))=(s/(ln (√2)))  ⇒(1/(2x))=((ln (√2))/s)  similarly  ⇒(1/(3y))=((ln (3)^(1/3) )/s)  ⇒(1/(5z))=((ln (5)^(1/5) )/s)  now we only need to compare the  numbers (√2), (3)^(1/3) , (5)^(1/5) .  3^2 =9>8=2^3   ⇒3>((√2))^3   ⇒(3)^(1/3) >(√2)  2^5 =32>25=5^2   ⇒2>((5)^(1/5) )^2   ⇒(√2)>(5)^(1/5)     ⇒(3)^(1/3) >(√2)>(5)^(1/5)   ⇒(1/(3y))>(1/(2x))>(1/(5z))  ⇒3y<2x<5z
$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} ={t}>\mathrm{0} \\ $$$${x}\:\mathrm{ln}\:\mathrm{2}={y}\:\mathrm{ln}\:\mathrm{3}={z}\:\mathrm{ln}\:\mathrm{5}=\mathrm{ln}\:{t}={s} \\ $$$$\Rightarrow{x}=\frac{{s}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{x}=\frac{{s}}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}}=\frac{{s}}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}=\frac{\mathrm{ln}\:\sqrt{\mathrm{2}}}{{s}} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{y}}=\frac{\mathrm{ln}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}}{{s}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{5}{z}}=\frac{\mathrm{ln}\:\sqrt[{\mathrm{5}}]{\mathrm{5}}}{{s}} \\ $$$${now}\:{we}\:{only}\:{need}\:{to}\:{compare}\:{the} \\ $$$${numbers}\:\sqrt{\mathrm{2}},\:\sqrt[{\mathrm{3}}]{\mathrm{3}},\:\sqrt[{\mathrm{5}}]{\mathrm{5}}. \\ $$$$\mathrm{3}^{\mathrm{2}} =\mathrm{9}>\mathrm{8}=\mathrm{2}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}>\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{5}} =\mathrm{32}>\mathrm{25}=\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}>\left(\sqrt[{\mathrm{5}}]{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}>\sqrt[{\mathrm{5}}]{\mathrm{5}} \\ $$$$ \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{3}}>\sqrt{\mathrm{2}}>\sqrt[{\mathrm{5}}]{\mathrm{5}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{y}}>\frac{\mathrm{1}}{\mathrm{2}{x}}>\frac{\mathrm{1}}{\mathrm{5}{z}} \\ $$$$\Rightarrow\mathrm{3}{y}<\mathrm{2}{x}<\mathrm{5}{z} \\ $$

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