UPSC NDA I 2024 Mathematics Paper

Concept:The equation is solved using the base change property $\log_{b^n}a = \frac{1}{n}\log_b a$, and the solution must satisfy the domain conditions: $x-1>0$ and $x-3>0$.Explanation:Rewrite $\log_4(x-1)$ as $\log_{2^2}(x-1) = \frac{1}{2}\log_2(x-1)$.The equation becomes $\frac{1}{2}\log_2(x-1) = \log_2(x-3)$.Multiply both sides by 2: $\log_2(x-1) = 2\log_2(x-3)$.Using $\log a^m=m\log a$, we get $\log_2(x-1) = \log_2((x-3)^2)$.Since logarithms are one-to-one, equate arguments: $x-1 = (x-3)^2$.Expand: $x-1 = x^2 - 6x + 9$.Rearrange: $0 = x^2 - 7x + 10$.Factor: $(x-5)(x-2)=0$.So potential solutions: $x=5$ and $x=2$.Now check domain: $\log_2(x-3)$ requires $x-3>0 \Rightarrow x>3$; $\log_4(x-1)$ requires $x-1>0 \Rightarrow x>1$.$x=2$ fails because $x-3=-1<0$ (logarithm undefined).$x=5$ satisfies both conditions.Thus only one valid solution exists.Answer:One (Option B).