For n=5=>25>5(5)=>32>25n=5=>25>5(5)=>32>25 true

For n=6=>26>5(6)=>64>30n=6=>26>5(6)=>64>30 true

For n=7=>27>5(7)=>128>35n=7=>27>5(7)=>128>35 true

We assume that for n=kn=k:

2k>5k…(1)2k>5k…(1)

We will show that for n=k+1n=k+1:

2k+1>5(k+1)…(2)2k+1>5(k+1)…(2)

We multiply both sides of (1)(1) by 22 and we take:

2k+1>10k…(3)2k+1>10k…(3)

Now if we show that 10k>5(k+1)10k>5(k+1), by the transitive property, we are done.

10k>5(k+1)=>5k>510k>5(k+1)=>5k>5

which is true, since k>5k>5.

Second proof:

We consider the function f(x)=2x−5x…(1)f(x)=2x−5x…(1)

By differentiating (1)(1), we obtain:

(2x−5x)′=>2x[ln(2)]−5(2x−5x)′=>2x[ln(2)]−5

We see that:

(2x)ln(2)−5>0=>x>[ln(5)−ln(ln(2))]/ln(2)=>x>2.85(2x)ln(2)−5>0=>x>[ln(5)−ln(ln(2))]/ln(2)=>x>2.85

Hence, since we care about integral values of xx, we conclude that for x≥3x≥3 our function is genuinely increasing function. Therefore, since we see that:

25−5(5)=725−5(5)=7, then, for every n>5n>5, we must have:

2n−5n>7=>2n−5n>0=>2n>5n2n−5n>7=>2n−5n>0=>2n>5n

Hence, the second proof is now complete.