# log 54 to the base 3/log 3 to the base 486-log 1458 to the base 3/log 3 to the base 18=xFind the value of x.

Arun Kumar IIT Delhi
10 years ago
Hello Student,
$\\a=log_{24}54={ln54 \over ln24}={ln3+ln18 \over ln2+ln12} \\a={3ln3+ln2 \over 3ln2+ln3}={3x+y \over 3y+x} \\b=log_{12}18={ln18 \over ln12} \\={2ln3+ln2 \over 2ln2+ln3}={2x+y \over 2y+x} \\=>ab-5(a-b) \\=>{3x+y \over 3y+x}*{2x+y \over 2y+x}-5({3x+y \over 3y+x}-{2x+y \over 2y+x}) \\=>{y^2+5xy+6x^2-5((3x^2+2y^2+7xy)-(3y^2+2x^2+7xy)) \over (2y+x)*(3y+x)} \\=>{y^2+5xy+6x^2 - 5(x^2-y^2) \over (2y+x)*(3y+x)} \\=>{6y^2+5xy+x^2 \over 6y^2+x^2+5xy} \\=>1$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
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