IPMAT Indore 2024 · Quantitative Ability · Question 18

Step 1 — convert to a common base. Write every log in base 10 (any fixed base works).

a={\log}_{4}x=\frac{\log x}{\log 4}=\frac{\log x}{2\log 2}\Rightarrow \log 2=\frac{\log x}{2a}

b={\log}_{25}x=\frac{\log x}{\log 25}=\frac{\log x}{2\log 5}\Rightarrow \log 5=\frac{\log x}{2b}

Step 2 — build \log 10. Since 10=2\times 5,

\log 10=\log 2+\log 5=\frac{\log x}{2a}+\frac{\log x}{2b}=\log x\cdot \frac{a+b}{2ab}

Step 3 — form the required log.

{\log}_{x}10=\frac{\log 10}{\log x}=\frac{a+b}{2ab}

Sanity check. Put x=100: a={\log}_{4}100\approx 3.322, b={\log}_{25}100\approx 1.431. Then \frac{a+b}{2ab}\approx \frac{4.753}{9.508}\approx 0.5, and indeed {\log}_{100}10=0.5 ✔

Answer: Option 3 — \frac{a+b}{2ab}

Shortcut. Flip to reciprocals. \frac{1}{a}={\log}_{x}4=2{\log}_{x}2 and \frac{1}{b}={\log}_{x}25=2{\log}_{x}5. Adding,

\frac{1}{a}+\frac{1}{b}=2\left({\log}_{x}2+{\log}_{x}5 \right)=2{\log}_{x}10

So {\log}_{x}10=\frac{1}{2}\left( \frac{1}{a}+\frac{1}{b}\right)=\frac{a+b}{2ab} in one line. Reciprocals of logs are the fast lane whenever the unknown has x as its base.