
Concept explainers
a.
To find:
The time required for a $1000 investment todouble at interest rate r, compounded continuously.Round your answer to two decimal places.
a.

Answer to Problem 144E
It will take approximately 11.55 years for the investment to double.
Explanation of Solution
Given:
r=6% .
Concept used:
Continuous compound interest formula- A=Pert , where,
A = Final amount after t years,
P = Principal amount,
r = Annual interest rate in decimal form.
Calculation:
Upon substituting our given values in continuous compound interest formula, we will get:
A=Pert(Continuous compound interest formula)2000=1000e0.06t(Substituting r=0.06)20001000=1000e0.06t10002=e0.06te0.06t=2(Switching sides)ln(e0.06t)=ln(2)(Taking natural log on both sides)0.06tln(e)=ln(2)(Using property ln(xm)=mln(x))0.06t=ln(2)t=ln(2)0.06t=11.552453009t≈11.55(Rounding to two deimals)
Therefore,it will take approximately 11.55 years for the investment to double.
b.
To find:
The time required for a $1000 investment to triple at interest rate r, compounded continuously.Round your answer to two decimal places.
b.

Answer to Problem 144E
It will take approximately 18.31 years for the investment to triple.
Explanation of Solution
Given:
r=6% .
Concept used:
Continuous compound interest formula- A=Pert , where,
A = Final amount after t years,
P = Principal amount,
r = Annual interest rate in decimal form.
Calculation:
Upon substituting our given values in continuous compound interest formula, we will get:
A=Pert(Continuous compound interest formula)3000=1000e0.06t(Substituting r=0.06)30001000=1000e0.06t10003=e0.06te0.06t=3(Switching sides)ln(e0.06t)=ln(3)(Taking natural log on both sides)0.06tln(e)=ln(3)(Using property ln(xm)=mln(x))0.06t=ln(3)t=ln(3)0.06t=18.3102048t≈18.31(Rounding to two deimals)
Therefore, it will take approximately 18.31 years for the investment to triple.
Chapter 3 Solutions
Precalculus with Limits: A Graphing Approach
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