IPMAT Indore 2023 · Quantitative Ability · Question 6

Step 1 — simplify the first term. Since {\log}_{2}\left({2}^{A}\right)=A for any A:

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

Step 2 — substitute. Put t={\log}_{2}x. The equation becomes a plain quadratic:

{t}^{2}-5t+6=0 \Rightarrow \left( t-2 \right)\left( t-3 \right)=0 \Rightarrow t=2, 3

Step 3 — return to x.

{\log}_{2}x=2\Rightarrow x=4,{\log}_{2}x=3\Rightarrow x=8

Both are positive, so both are in the domain. ✔

Step 4 — multiply.

4\times 8=32

Answer: 32

Shortcut. Use the sum of the roots in t, not the roots themselves.

{x}_{1}{x}_{2}={2}^{{t}_{1}}\cdot{2}^{{t}_{2}}={2}^{\,{t}_{1}+{t}_{2}}

and for {t}^{2}-5t+6=0 the sum of roots is 5 by Vieta — no factorising needed. So

\text{product}={2}^{5}=32

This generalises: whenever a log substitution turns an equation into a quadratic in t={\log}_{b}x, the product of the x-roots is {b}^{\,\left( \text{sum of}t\text{-roots}\right)}. Vieta gives you that sum directly from the coefficients.