The problem thus reduces to finding the least value of n such that
(109)n<xn−4≤2201 and (1110)n−1>xn−1−4>2201
Taking logarithms, we get nln109<−20ln2 and (n−1)ln1110>−20ln2, i.e.
n>ln91020ln2 and n−1<ln101120ln2
As approximations, we can use ln910≈91,ln1011≈101, and ln2≈0.7. These approximations allow us to estimate
126<n<141
which gives (C)[81,242].
OR
The condition where xm≤4+2201 gives the motivation to make a substitution to change the equilibrium from 4 to 0 . We can substitute xn=yn+4 to achieve that. Now, we need to find the smallest value of m such that ym≤2201 given that y0=1.
Factoring the recursion xn+1=xn+6xn2+5xn+4, we get:
Using wishful thinking, we can simplify the recursion as follows:
yn+1=yn+10yn2+9yn+yn−yn
yn+1=yn+10yn(yn+10)−yn
yn+1=yn−yn+10yn
yn+1=yn(1−yn+101)
The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the yn sequence is strictly decreasing, so all the terms after y0 will be less than 1 . Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.
With both of those observations in mind, 109<1−yn+101≤1110. Combining this with the fact that the recursion resembles a geometric sequence, we conclude that (109)n<yn≤(1110)n.
109 is approximately equal to 1110 and the ranges that the answer choices give us are generous, so we should use either 109 or 1110 to find a rough estimate for m.
Since 21=0.5, that means 21=2−21≈0.7. Additionally, (109)3=0.729
Therefore, we can estimate that 2−21<y3.
Raising both sides to the 40th power, we get 2^{-20}<\left(y_{3}\right)^
But y3=(y0)3, so 2−20<(y0)120 and therefore, 2−20<y120.
This tells us that m is somewhere around 120, so our answer is (C)[81,242].