0JEE 2011 Question on Logarithm
Let (x0,y0) be the solution of the following equations: (2x)ln2 = (3y)ln3 3lnx = 2lny
Then x0 is
(A)
1 6
(B)
1 3
(C)
1 2
(D) 6
Answer and Comments: (C). We can treat the data as a system of equations in the unknowns lnx and lny instead of x and y. This is permissible because a (positive) real number is uniquely determined by its logarithm. So, taking logarithms, we get
(ln2)2 + (ln2)(lnx) = (ln3)2 + (ln3)(lny) (1) (ln3)(lnx) = (ln2)(lny) (2)
which can be treated as a system of linear equations in the unknowns lnx and lny. Eliminating lny from the two equations gives (ln2)3 + (ln2)2(lnx)−(ln3)2(ln2) = (ln3)2(lnx) (3)
5
which simpliï¬?es to [(ln3)2 −(ln2)2](lnx) =−(ln2)[(ln3)2 −(ln2)2] (4) As (ln3)6= ±(ln2), we cancel the bracketed factor and get lnx = −ln2 (5) which gives x = 1/2. Once the key idea strikes, viz. to take lnx and lny as the variables, the problem is straightforward. The properties of logarithms needed are elementary and standard. Nowhere we have to convert logarithms w.r.t. one base to those w.r.t. another.
