IPMAT Indore 2025 · Quantitative Ability · Question 17

Step 1 — isolate the log term.

y-a=b{\log}_{e}x

Step 2 — exponentiate both sides. Raise e to the power of each side of the original equation:

{e}^{y}={e}^{\,a+b{\log}_{e}x}={e}^{a}\cdot{e}^{\,b{\log}_{e}x}

Step 3 — simplify the second factor. Using b{\log}_{e}x={\log}_{e}{x}^{b} and {e}^{{\log}_{e}M}=M:

{e}^{\,b{\log}_{e}x}={x}^{b}

So

{e}^{y}=\underset{\text{a constant}}{\underset{⏟}{{e}^{a}}}\cdot{x}^{b}

Since {e}^{a} is a fixed number, {e}^{y} is directly proportional to {x}^{b}.

Why the others fail: y-a=b{\log}_{e}x is proportional to {\log}_{e}x, not to {x}^{b} — which rules out options 1 and 2. Option 4 would require {\log}_{e}y to be a constant multiple of x, which it is not.

Answer: Option 3 — {e}^{y} is proportional to {x}^{b}