Question

Solve this application using logarithms.

Answer 1

Part A:

Given that a man invested $2,000 in savings at 6%. In 19 years, the value of the investment will be:

$FV=2000(1+0.06)^{19} \\ \\ =2000(1.06)^{19}=2000(3.0256) \\ \\ \approx\ 6,051$



Part B:

Given that a man invested $2,000 in a mutual fund earning 6.5%. In 19 years, the value of the investment will be:

$FV=2000(1+0.065)^{19} \\ \\ =2000(1.065)^{19}=2000(3.3086) \\ \\ \approx\ 6,617$



Part C:

Given that a man invested $2,000 in savings at 6%. The number of years it will take for the value of the investment to double is obtained as follows:

Let the required number of years be n, then

$2000(1.06)^n=4000 \\ \\ \Rightarrow(1.06)^n=2 \\ \\ \Rightarrow n\log{(1.06)}=\log{2} \\ \\ \Rightarrow n= \frac{\log{2}}{\log{1.06}} = \frac{0.6931}{0.0583} =11.9\approx12$

Therefore, it will take the savings 12 years to double in value.



Part D:

Given that a man invested $2,000 in a mutual fund earning 6.5%. The number of years it will take for the value of the investment to double is obtained as follows:

Let the required number of years be n, then

$2000(1.065)^n=4000 \\ \\ \Rightarrow(1.065)^n=2 \\ \\ \Rightarrow n\log{(1.065)}=\log{2} \\ \\ \Rightarrow n= \frac{\log{2}}{\log{1.065}} = \frac{0.6931}{0.0630}\approx11$

Therefore, it will take the investment 11 years to double in value.