IPMAT Indore 2024 · Quantitative Ability · Question 23

Step 1 — abbreviate. Let p={\log}_{7}2 and q={\log}_{7}5. Then

{\log}_{7}4=2p,{\log}_{7}8=3p,{\log}_{7}25=2q

Step 2 — substitute.

a=\frac{\left( 2p \right)\left( q-p \right)}{\left( 2q \right)\left( 3p-2p \right)}=\frac{2p\left( q-p \right)}{2qp}=\frac{q-p}{q}

Step 3 — simplify.

a=1-\frac{p}{q}=1-\frac{{\log}_{7}2}{{\log}_{7}5}=1-{\log}_{5}2

Step 4 — exponentiate.

{5}^{a}={5}^{\,1-{\log}_{5}2}=\frac{{5}^{1}}{{5}^{{\log}_{5}2}}=\frac{5}{2}

using {5}^{{\log}_{5}2}=2.

Answer: Option 4 — \frac{5}{2}

Shortcut. Collapse the differences first. {\log}_{7}5-{\log}_{7}2={\log}_{7}5/2 and {\log}_{7}8-{\log}_{7}4={\log}_{7}2. So

a=\frac{{\log}_{7}4\cdot{\log}_{7}\frac{5}{2}}{{\log}_{7}25\cdot{\log}_{7}2}=\frac{2{\log}_{7}2\cdot{\log}_{7}\frac{5}{2}}{2{\log}_{7}5\cdot{\log}_{7}2}=\frac{{\log}_{7}\frac{5}{2}}{{\log}_{7}5}={\log}_{5}\frac{5}{2}

Therefore {5}^{a}=\frac{5}{2} — the answer is sitting inside the logarithm. Turning differences into quotients before anything else saves the whole middle of the solution.