Solve this application using logarithms. At his son's birth, a man invested $2,000 in savings at 6% for his son's college education. Approximately how much, to the nearest dollar, will be available in 19 years?≈$At his son's birth, a man invested $2,000 in a mutual fund earning 6.5% for his son's college education. Approximately how much, to the nearest dollar, will be available in 19 years?real value ≈ $For each of the man’s investments described above, how many years will it take for each investment to double in value? Round your answers to the nearest whole year.years to double savings accountyears to double mutual fund

Part A:

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

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



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:

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



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

[tex]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[/tex]

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

[tex]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[/tex]

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