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MISCELLANEOUS EXERCISE 6 (II) · Q181

Q.If log⁡ax+y−2z=log⁡by+z−2x=log⁡cz+x−2y\dfrac{\log a}{x+y-2z}=\dfrac{\log b}{y+z-2x}=\dfrac{\log c}{z+x-2y}, show that abc=1abc=1.

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Let log⁡ax+y−2z=log⁡by+z−2x=log⁡cz+x−2y=k\dfrac{\log a}{x+y-2z}=\dfrac{\log b}{y+z-2x}=\dfrac{\log c}{z+x-2y}=k.

Then log⁡a=k(x+y−2z)\log a=k(x+y-2z), log⁡b=k(y+z−2x)\log b=k(y+z-2x), log⁡c=k(z+x−2y)\log c=k(z+x-2y).

Add all three: log⁡a+log⁡b+log⁡c=k[(x+y−2z)+(y+z−2x)+(z+x−2y)]\log a+\log b+\log c=k\big[(x+y-2z)+(y+z-2x)+(z+x-2y)\big].

Expand the bracket: x+y−2z+y+z−2x+z+x−2y=(x−2x+x)+(y+y−2y)+(−2z+z+z)=0+0+0=0x+y-2z+y+z-2x+z+x-2y=(x-2x+x)+(y+y-2y)+(-2z+z+z)=0+0+0=0. …

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