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Question Number 101225 by harckinwunmy last updated on 01/Jul/20

Commented by Dwaipayan Shikari last updated on 01/Jul/20

x^a =y^b =z^c =k(k≠0) x=k^(1/a) y=k^(1/b) z=k^(1/c) y^2 =k^(2/b) xz=k^((1/a)+(1/c)) (2/b)=(1/a)+(1/c)⇒((2a−b)/(ab))=(1/c)⇒c=((ab)/(2a−b))

$${x}^{{a}} ={y}^{{b}} ={z}^{{c}} ={k}\left({k}\neq\mathrm{0}\right) \\ $$$${x}={k}^{\frac{\mathrm{1}}{{a}}} \:{y}={k}^{\frac{\mathrm{1}}{{b}}} \:\:\:{z}={k}^{\frac{\mathrm{1}}{{c}}} \\ $$$${y}^{\mathrm{2}} ={k}^{\frac{\mathrm{2}}{{b}}} \:\:{xz}={k}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}} \\ $$$$\frac{\mathrm{2}}{{b}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}\Rightarrow\frac{\mathrm{2}{a}−{b}}{{ab}}=\frac{\mathrm{1}}{{c}}\Rightarrow{c}=\frac{{ab}}{\mathrm{2}{a}−{b}} \\ $$$$ \\ $$

Commented by smridha last updated on 01/Jul/20

2.((log(x))/((y−z)))=((log(y))/((z−x)))=((log(z))/((x−y)))=u now x^x .y^y .z^z =e^(xlog(x)+ylog(y)+zlog(z)) putting the values we get =e^(ux(y−z)+uy(z−x)+uz(x−y)) =e^0 =1

$$\mathrm{2}.\frac{\boldsymbol{{log}}\left(\boldsymbol{{x}}\right)}{\left(\boldsymbol{{y}}−\boldsymbol{{z}}\right)}=\frac{\boldsymbol{{log}}\left(\boldsymbol{{y}}\right)}{\left(\boldsymbol{{z}}−\boldsymbol{{x}}\right)}=\frac{\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)}{\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)}=\boldsymbol{{u}} \\ $$$$\boldsymbol{{now}}\: \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{x}}} .\boldsymbol{{y}}^{\boldsymbol{{y}}} .\boldsymbol{{z}}^{\boldsymbol{{z}}} =\boldsymbol{{e}}^{\boldsymbol{{xlog}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{ylog}}\left(\boldsymbol{{y}}\right)+\boldsymbol{{zlog}}\left(\boldsymbol{{z}}\right)} \\ $$$$\boldsymbol{{putting}}\:\boldsymbol{{the}}\:\boldsymbol{{values}}\:\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\:\:\:\:\:\:\:=\boldsymbol{{e}}^{\boldsymbol{{ux}}\left(\boldsymbol{{y}}−\boldsymbol{{z}}\right)+\boldsymbol{{uy}}\left(\boldsymbol{{z}}−\boldsymbol{{x}}\right)+\boldsymbol{{uz}}\left(\boldsymbol{{x}}−\boldsymbol{{y}}\right)} \\ $$$$\:\:\:\:\:\:\:=\boldsymbol{{e}}^{\mathrm{0}} =\mathrm{1} \\ $$

Commented by bemath last updated on 01/Jul/20

(3a)(√(p+(√q))) = (√x) + (√y) (√(((√x)+(√y))^2 )) = (√(x+y+2(√(xy)))) → { ((p=x+y)),((q=4xy)) :}

$$\left(\mathrm{3a}\right)\sqrt{\mathrm{p}+\sqrt{\mathrm{q}}}\:=\:\sqrt{\mathrm{x}}\:+\:\sqrt{\mathrm{y}} \\ $$$$\sqrt{\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}} \\ $$$$\rightarrow\begin{cases}{\mathrm{p}=\mathrm{x}+\mathrm{y}}\\{\mathrm{q}=\mathrm{4xy}}\end{cases}\: \\ $$

Commented by Dwaipayan Shikari last updated on 01/Jul/20

p+(√q)=x+y+2(√(xy)) p=x+y (√q)=2(√(xy)) (p^2 −q)^(1/2) =((x+y)^2 −4xy)^(1/2) =x−y

$${p}+\sqrt{{q}}={x}+{y}+\mathrm{2}\sqrt{{xy}} \\ $$$${p}={x}+{y} \\ $$$$\sqrt{{q}}=\mathrm{2}\sqrt{{xy}} \\ $$$$\left({p}^{\mathrm{2}} −{q}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={x}−{y} \\ $$$$ \\ $$

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