IPMAT Indore 2022 · Quantitative Ability · Question 1

Step 1 — reduce both logs to one quantity. Put L={\log}_{x}y. Using the change-of-base rule,

{\log}_{{x}^{2}}y=\frac{\log y}{2\log x}=\frac{L}{2},{\log}_{{y}^{2}}x=\frac{\log x}{2\log y}=\frac{1}{2L}

Step 2 — solve the resulting equation.

\frac{L}{2}+\frac{1}{2L}=1 \Rightarrow{L}^{2}+1=2L \Rightarrow{\left( L-1 \right)}^{2}=0

So L=1, a repeated root — which means {\log}_{x}y=1, i.e.

y=x

Step 3 — use the second condition.

x={x}^{2}-30 \Rightarrow{x}^{2}-x-30=0 \Rightarrow \left( x-6 \right)\left( x+5 \right)=0

The base of a logarithm must be positive and \ne 1, so x=6 and y=6.

Step 4 — evaluate.

{x}^{2}+{y}^{2}=36+36=72

Answer: 72

Shortcut. Spot the AM–GM shape. For any positive L,

\frac{L}{2}+\frac{1}{2L}\ge 2\sqrt{\frac{L}{2}\cdot \frac{1}{2L}}=1

with equality only when the two terms are equal, i.e. L=1. So the equation “=1” is the equality case of AM–GM, forcing y=x instantly — no quadratic needed.

Then x={x}^{2}-30 gives x=6, and the answer is 2\times 36=72. Whenever you see t/2+1/2t=1 or t+1/t=2, read off t=1 immediately.