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x(y+z-x)/logx = y(x+z-y)/logy = z(x+y-z)/logz so show that, x y y x = y z z y = z x x z ................

x(y+z-x)/logx = y(x+z-y)/logy = z(x+y-z)/logz  so show that,


 


                  xyyx = yzzy = zxxz ................Laughing

Grade:12

1 Answers

Arpit Jaiswal
31 Points
11 years ago

nice question but cracked  Wink

let the given be equal to k-

x(y+z-x)/logx = y(x+z-y)/logy = z(x+y-z)/logz    =k

 

or, logx=x(y+z-x)/k, multiplying by y on both sides, we get-   ylogx=xy(y+z-x)/k      ....(1)

also, multiplying both sides by z we get-     zlogx=zx(y+z-x)/k   ........(2)

 

and, logy=y(x+z-y)/k, multiplying both sides by x, we get-      xlogy=xy(x+z-y)/k    .......(3)

also, multiplying both sides by z we get-     zlogy=zy(x+z-y)/k   ...........(4)

 

and, logz=z(x+y-z)/k, multipying both sides by y, we get-        ylogz=yz(x+y-z)/k    .......(5)

also, mulitplying both sides by x we get-     xlogz=xz(x+y-z)/k    ..........(6)

 

 

adding 1 and 3, we have- log(x^y)(y^x)=xyz/k

adding 4 and 5, we have- log(y^z)(z^y)=xyz/k

adding 2 and 6, we have- log(z^x)(x^z)=xyz/k

comparaing the RHS of above and removing log, we have-

(x^y)(y^x)=(y^z)(z^y)=(x^z)(z^x)

Hence proved.Laughing

AJ

 
 


 

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