Question

A savings account earns 2.5% interest. If $3000 is invested, how many years is it until each of the following amounts is on deposit?

Where F is future, P is principal, F/P is Compound Amount Factor ("find F given P"), and i is interest percentage. If possible, use these equations to help solve for the answers: F = P(1+i)ⁿ  $\Rightarrow$F/P = (1+i)ⁿ;

log (F/P) = log [(1+i)ⁿ = nlog (1+i) $\Rightarrow$ n = log(F/P) / log (1+i).

Please show all steps and do not use spreadsheet to solve.

a) $3394

b) $3655

c) $4035

d) $5165

Answer 1

Interest note (i) = 2.5% or 0.025 principal amount (p) = $3000. (a) Future amount (f) = $3394 f = p (r + i)^n 3394 = 3000 (1 + 0.025)^n (1.025)^n = 3394/3000 (1.025)^n = 1.13 Taking log both sides. N log (1.025) = log 1.13 n = log (1.13)/log (1.025) n = log (F/P)/log (1 + i) n = 4.9 n = 5 years (b) F = $ 3655. F = P (1 + i)^n n = log (F/P)/log (1 + i) n = log (3655/3000)/log (1 + 0.025) n = log (1.21)/log (1.025)n = 7.7 years \r\n (c) F = $ 4035 F = P (1 + i)^n n = log (F/P)/log (1 + i) n = log (4035/3000)/log (1 + 0.025) n = log (1.345)/log (1.025) n = 12 rightarrow n = 12 years (d) f = $ 5165 f = p (1 + i)^n n = log (F + P)/log (1 + i) n = log