Compound interest

Question

At an effective annual compound interest of i>0, it is found that an investment doubles in a years, triples in b years and 1 grows to 5 in c years.

1. Is c greater than (3a+b)/2?
2. To what amount does 10 grow in (3a + 2b) years?
3. In how many years will 1000 grow to 1200?

Answer 1

2=e^ai

ai=ln2

a=ln2/i

3=ebi

b= ln3/i

c=ln5/i

3a+b= (3ln2+ln3)/i= ln24/i

(3a+b)/2 = ln24/2i

ln5>ln24/2, 1.6>1.58

ln5/i>ln24/2i

c> (a+b)/2