IPMAT Indore 2023 · Quantitative Ability · Question 17

Step 1 — unwind each condition. {\log}_{n}\left({\log}_{2}a \right)=1 means {\log}_{2}a={n}^{1}, hence

a={2}^{n}, b={2}^{{n}^{2}}, c={2}^{{n}^{3}}

Step 2 — notice the chain. Each exponent is the previous one multiplied by n:

{a}^{n}={\left({2}^{n}\right)}^{n}={2}^{{n}^{2}}=b,{b}^{n}={2}^{{n}^{3}}=c

So the three numbers satisfy b={a}^{n} and c={b}^{n}.

Step 3 — test option (a). Using {a}^{n}=b:

{\left({a}^{n}+b \right)}^{n}={\left( b+b \right)}^{n}={\left( 2b \right)}^{n}={2}^{n}\,{b}^{n}={2}^{n}\cdot c

And the right-hand side:

ac={2}^{n}\cdot c

They are equal. ✔

Step 4 — dismiss the rest.

(b) {a}^{n}+{b}^{n}=b+c, which is not {c}^{n} in general.

(c) a+b=c would need {2}^{n}+{2}^{{n}^{2}}={2}^{{n}^{3}} — false for, say, n=2 (4+16\ne 256).

(d) {\left( b-a \right)}^{n}=c-b fails at n=2: {\left( 16-4 \right)}^{2}=144 but 256-16=240.

Answer: Option 1 — {\left({a}^{n}+b \right)}^{n}=ac

Shortcut. Put n=2 and the whole question becomes arithmetic. Then

a={2}^{2}=4, b={2}^{4}=16, c={2}^{8}=256

Now test each option in one line:

Substituting a convenient value is almost always faster than algebra when all four options are identities to be checked. Pick the smallest legal n and compute.