IIT JEE , AIEEE Forum - Mathematics , Algebra

apoorva_43

If 3 , 3log_yx , 3log_zy , 7log_xz are in AP then

1)x¹⁸=y²¹=z¹⁴

2)x¹⁸=y²¹=z²⁸

3)x⁶=y²¹=z²⁸

4)x¹⁸=y⁷=z¹⁴

ramkumar_november

the answer is x¹⁸=y²¹=z²⁸   . the proof is .........
let log_yx=a, log_zy=b, log_xz=c .......
given 3,3a,3b,7c are in AP..
6a=3+3b => 2a=b+1..........(i)
6b=3a+7c................(ii)
abc=1.........(iii)
multiply (ii) by ab we get
6ab²=3a²b+7abc
6a(2a-1)²=3a²(2a-1)+7
solving this equation we get a=7/6
substitute a=7/6 in (i) we get b=4/3
substitute values of a and b in (iii) we get c=9/14
therefore
log_yx=7/6 => x=y^7/6 => x⁶=y⁷ => x¹⁸=y²¹
log_zy=4/3 => y=z^4/3 =>y³=z⁴   => y²¹=z²⁸
log_xz=9/14 => z=x^9/14 => z¹⁴=x⁹ => z²⁸=x¹⁸

thus we get   x¹⁸=y²¹=z²⁸   ......
answer 2) is correct

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