IPMAT Indore 2024 · Quantitative Ability · Question 9

Step 1 — clear the exponentials. Use {a}^{{\log}_{b}c}={c}^{{\log}_{b}a}, or more directly:

{4}^{{\log}_{2}x}={\left({2}^{2}\right)}^{{\log}_{2}x}={2}^{\,2{\log}_{2}x}={\left({2}^{{\log}_{2}x}\right)}^{2}={x}^{2}

{9}^{{\log}_{3}y}={\left({3}^{2}\right)}^{{\log}_{3}y}={\left({3}^{{\log}_{3}y}\right)}^{2}={y}^{2}

Step 2 — rewrite the equation.

{x}^{2}-4x+{y}^{2}-16y+68=0

Step 3 — complete both squares.

\left({x}^{2}-4x+4 \right)+\left({y}^{2}-16y+64 \right)=4+64-68=0

{\left( x-2 \right)}^{2}+{\left( y-8 \right)}^{2}=0

Step 4 — a sum of two squares is zero only if both are zero.

x=2, y=8 \Rightarrow y-x=6

Both are in the domain of the logarithms (x,y>0) ✔

Answer: 6

Shortcut. Two reflexes. {4}^{{\log}_{2}x}={x}^{2} and {9}^{{\log}_{3}y}={y}^{2} — whenever the outer base is the square of the log base, you get the square of the argument. After that, a two-variable quadratic equal to zero with no cross term is always a sum of squares: force each bracket to vanish. x=2, y=8, so y-x=6.