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Question Number 79978 by mr W last updated on 29/Jan/20

$$Givenforx,y,z>0:$$ $$2^x=3^y=5^z$$ $$Arrange2x,3y,5zinincreasingorder.$$

Answered by mind is power last updated on 29/Jan/20

$$xln\left(2\right)=yln\left(3\right)=zln\left(5\right)$$ $$y=\frac{xln\left(2\right)}{ln\left(3\right)}$$ $$z=\frac{xln\left(2\right)}{ln\left(5\right)}$$ $$5z=\frac{5}{ln\left(5\right)}.ln\left(2\right)x$$ $$3y=\frac{3ln\left(2\right)}{ln\left(3\right)}x$$ $$ln\left(8\right)<ln\left(9\right)\Rightarrow$$ $$3ln\left(2\right)<2ln\left(3\right)$$ $$\Rightarrow \frac{3ln\left(2\right)}{ln\left(3\right)}<2$$ $$\Rightarrow 3y=\frac{3ln\left(2\right)}{ln\left(3\right)}x<2x$$ $$\frac{5ln\left(2\right)}{ln\left(5\right)}>2\iff ln\left(32\right)>ln\left(25\right)True\Rightarrow$$ $$5z=\frac{5ln\left(2\right)}{ln\left(5\right)}x>2x\Rightarrow 5z>2x>3y$$

Answered by key of knowledge last updated on 29/Jan/20

$$2^x=3^y=5^z\Rightarrow \mathrm{x}.{\log }_{3}2=\mathrm{y}=\mathrm{z}.{\log }_{3}5$$ $$2\mathrm{x}=2\left(\frac{\mathrm{y}}{{1\mathrm{og}}_{3}2}\right)=\mathrm{y}.{2\log }_{2}3\approx 3.16\mathrm{y}$$ $$5\mathrm{z}=...=\mathrm{z}\times {5\log }_{5}3\approx 3.41\mathrm{y}$$ $$\Rightarrow 3\mathrm{y}<2\mathrm{x}<5\mathrm{z}$$

Answered by mr W last updated on 30/Jan/20

$$2^x=3^y=5^z=t>0$$ $$x\ln 2=y\ln 3=z\ln 5=\ln t=s$$ $$\Rightarrow x=\frac{s}{\ln 2}$$ $$\Rightarrow 2x=\frac{s}{\frac{1}{2}\ln 2}=\frac{s}{\ln \sqrt{2}}$$ $$\Rightarrow \frac{1}{2x}=\frac{\ln \sqrt{2}}{s}$$ $$similarly$$ $$\Rightarrow \frac{1}{3y}=\frac{\ln \sqrt[3]{3}}{s}$$ $$\Rightarrow \frac{1}{5z}=\frac{\ln \sqrt[5]{5}}{s}$$ $$nowweonlyneedtocomparethe$$ $$numbers\sqrt{2},\sqrt[3]{3},\sqrt[5]{5}.$$ $$3^2=9>8=2^3$$ $$\Rightarrow 3>{\left(\sqrt{2}\right)}^3$$ $$\Rightarrow \sqrt[3]{3}>\sqrt{2}$$ $$2^5=32>25=5^2$$ $$\Rightarrow 2>{\left(\sqrt[5]{5}\right)}^2$$ $$\Rightarrow \sqrt{2}>\sqrt[5]{5}$$ $$$$ $$\Rightarrow \sqrt[3]{3}>\sqrt{2}>\sqrt[5]{5}$$ $$\Rightarrow \frac{1}{3y}>\frac{1}{2x}>\frac{1}{5z}$$ $$\Rightarrow 3y<2x<5z$$

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