The question looks like this: $1000^1001 (?) 1001^1000$ We want to find the (?) operator. I scribbled that using logarithmic , here it goes. Scribble $1000^1001 (?) 1001^1000$ Let us assume: $1000 = a$ Therefore: $1001 = (a + 1)$ So this: $1000^1001 (?) 1001^1000$ Becomes: $a^(a+1) (?) (a + 1)^a$ ➡️ We then place log (logarithmic function) on both sides. This form: $a^(a+1) (?) (a + 1)^a$ ➡️ Becomes: $log_a(a^(a+1)) (?) log_a((a + 1)^a)$ Continue by implementing logarithmic properties . The form above can be transformed into this: $(a + 1) log_a(a) (?) (a) log_a(a + 1)$ We can simplify $log_a(a) = 1$. Thus: $(a + 1) (1) (?) (a) log_a(a + 1)$ $log_a(a + 1) > 1$ — very near 1 because $a = 1000$. Therefore we can safely assume that $log_a(a + 1) = 1$ Final simplification: ➡️ $(a + 1) (1) (?) (a) (1)$ $(a + 1) (?) (a)$ For any number, adding one gives a result larger than the number itself — naturally . Thus: $(a + 1) > a$...