$2^40$, $3^21$, $4^18$ और $8^12$ में से कौन-सी संख्या लघुतम है?

Correct Option: 3^21

To find the smallest number among 2^40, 3^21, 4^18, and 8^12, let us express all numbers with a common exponent or base. Let's convert them to base 2 or simplify their exponents by finding the HCF of the exponents (40, 21, 18, 12), which is 3. Let's rewrite the expressions: 2^40 = (2^40) - let's check powers of 3: 3^21 = (3^7)^3 = (2187)^3. 4^18 = (4^6)^3 = (4096)^3. 8^12 = (8^4)^3 = (4096)^3. Now let's compare with exponent 3: 2^40 = (2^40/3) - instead, let's use base 2 for all: 2^40 = 2^40. 3^21 = (3^7)^3 = 2187^3. 4^18 = (2^2)^18 = 2^36. 8^12 = (2^3)^12 = 2^36. Notice that 4^18 = 2^36 and 8^12 = 2^36. Comparing remaining numbers: 2^36, 2^40, and 3^21. Since 3^21 = (3^3)^7 = 27^7, or let's look at approximate values: 3^21 is much smaller than 2^40 (since 2^40 is extremely large). Wait, let's compare 2^36 and 3^21: 2^12 = 4096, so 2^36 = (4096)^3. 3^3 = 27, so 3^21 = (27)^7. Clearly 27^7 is smaller than 4096^3. Wait, let's check 3^21 vs 2^36: 3^7 = 2187, while 2^12 = 4096. 2^36 = (2^12)^3 = 4096^3. 3^21 = (3^7)^3 = 2187^3. Since 2187 < 4096, 2187^3 < 4096^3, so 3^21 < 4^18 (or 8^12). What about 2^40? 2^40 = 2^4 x 2^36 = 16 x 2^36, which is larger than 2^36. Thus, 3^21 is the smallest number.