After "26 years" the amount in A2 become greater than that in A1.
According to the question,
The future value of funds in A1:
= [tex]100000(1.08)^n[/tex]
here, n = Number of years
The future value of A2 will be:
= [tex]\frac{10000}{0.08}[1.08^n-1][/tex]
= [tex]125000[1.08^n-1][/tex]
now,
→ [tex]125000[1.08^n -1] > 100000(1.08)^n[/tex]
→ [tex]25000(1.08^n) > 125000[/tex]
→ [tex]1.08^n > 5[/tex]
By taking "log", we get
→ [tex]n> \frac{ln \ 5}{ln \ 1.08} = 20.9[/tex] or, [tex]21 \ years[/tex] (n "Number of years")
Each year, with $7000 in A2
→ [tex]\frac{7000}{0.08} [1.08^n -1]> 100000(1.08^n)[/tex]
→ [tex]1.08^n -1 > 1.142857 (1.08^n)[/tex]
→ [tex]1.08^n> \frac{1}{0.142857} = 7[/tex]
By taking "log", we get
→ [tex]n > \frac{ln \ 7}{ln \ 1.08} = 25.28[/tex] or, [tex]26 \ years[/tex] (n "Number of years")
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