^You might be aware of the fact that the exponential functions grow much faster than polynomial functions.

Let us assume for now that **for all** values of n, e < d and e < c.

Now lets illustrate by counterexample that our assumption based on our intuition is **wrong!**

Consider some large value of n.

Let n = 2^32

c) (2^32)^1.75 = 7.2057594e+16

d) (2^32)(log 2^32)^9 = 3.0675921e+18 //base 10

e) 1.0000001^(2^32) = 3.373265e+186

Hence, our intuition based assumption that **for all** values of n, e < d and e < c is false.

Let's visualise these functions:

//Observe colors infront of the functions to find the corresponding curve in the graph

// Eg: Red <=> n^1.75

Above graph justifies our intuition that 1.0000001^n is approx 1^n i.e. is 1.

Lets zoom out a bit.

Still the curve of 1.0000001^n is hovering around 1.

Let's zoom out more.

Still nothing.

Voilla!

Observe how 1.0000001^n has stopped hovering around 1 and grown so quickly as n has become large.

As we keep on increasing n, at some point e) will overtake c) and d) as we have seen in our

calculation performed above.

PS: 1) Don't confuse between 1.0000001^32 and 1.0000001^(2^32) .

^ ^

| |

= 1 = 3.373265e+186

2) From above calculation as well as graphs, it appears that c < d, therefore** **option(c)

looks like more apt. answer.