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Question Number 113886 by Aina Samuel Temidayo last updated on 16/Sep/20

Answered by MJS_new last updated on 16/Sep/20

(a)

$$\left({a}\right) \\ $$

Commented by Aina Samuel Temidayo last updated on 16/Sep/20

Solution please?

$$\mathrm{Solution}\:\mathrm{please}? \\ $$

Commented by MJS_new last updated on 16/Sep/20

lol... just use your brain

$$\mathrm{lol}...\:\mathrm{just}\:\mathrm{use}\:\mathrm{your}\:\mathrm{brain} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Sep/20

a^x =(x+y+z)^y ,a^y =(x+y+z)^z ,a^z =(x+y+z)^x   a=(x+y+z)^(y/x) =(x+y+z)^(z/y) =(x+y+z)^(x/z)           (y/x)=(z/y)=(x/z)=((x+y+z)/(x+y+z))=1         y=x,z=y,x=z        x=y=z=k(say)  a^k =(k+k+k)^k   a=3k  k=a/3=x=y=z

$${a}^{{x}} =\left({x}+{y}+{z}\right)^{{y}} ,{a}^{{y}} =\left({x}+{y}+{z}\right)^{{z}} ,{a}^{{z}} =\left({x}+{y}+{z}\right)^{{x}} \\ $$$${a}=\left({x}+{y}+{z}\right)^{{y}/{x}} =\left({x}+{y}+{z}\right)^{{z}/{y}} =\left({x}+{y}+{z}\right)^{{x}/{z}} \\ $$$$\:\:\:\:\:\:\:\:\frac{{y}}{{x}}=\frac{{z}}{{y}}=\frac{{x}}{{z}}=\frac{{x}+{y}+{z}}{{x}+{y}+{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{y}={x},{z}={y},{x}={z} \\ $$$$\:\:\:\:\:\:{x}={y}={z}={k}\left({say}\right) \\ $$$${a}^{{k}} =\left({k}+{k}+{k}\right)^{{k}} \\ $$$${a}=\mathrm{3}{k} \\ $$$${k}={a}/\mathrm{3}={x}={y}={z} \\ $$

Answered by Rasheed.Sindhi last updated on 17/Sep/20

Multiplying all the three  a^(x+y+z) =(x+y+z)^(x+y+z)   a=x+y+z...................A  a^x =a^y ,a^y =a^z ,a^z =a^x   (a^x /a^y )=(a^y /a^z )=(a^z /a^x )=1  a^(x−y) =a^(y−z) =a^(z−x) =a^0   x−y=y−z=z−x=0  x=y=z=k(say)  In  A  a=k+k+k=3k⇒k=a/3  k=x=y=z=a/3

$${Multiplying}\:{all}\:{the}\:{three} \\ $$$${a}^{{x}+{y}+{z}} =\left({x}+{y}+{z}\right)^{{x}+{y}+{z}} \\ $$$${a}={x}+{y}+{z}...................\mathrm{A} \\ $$$${a}^{{x}} ={a}^{{y}} ,{a}^{{y}} ={a}^{{z}} ,{a}^{{z}} ={a}^{{x}} \\ $$$$\frac{{a}^{{x}} }{{a}^{{y}} }=\frac{{a}^{{y}} }{{a}^{{z}} }=\frac{{a}^{{z}} }{{a}^{{x}} }=\mathrm{1} \\ $$$${a}^{{x}−{y}} ={a}^{{y}−{z}} ={a}^{{z}−{x}} ={a}^{\mathrm{0}} \\ $$$${x}−{y}={y}−{z}={z}−{x}=\mathrm{0} \\ $$$${x}={y}={z}={k}\left({say}\right) \\ $$$${In}\:\:\mathrm{A} \\ $$$${a}={k}+{k}+{k}=\mathrm{3}{k}\Rightarrow{k}={a}/\mathrm{3} \\ $$$${k}={x}={y}={z}={a}/\mathrm{3} \\ $$

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