# Intro Calc HW

• Apr 3rd 2008, 03:05 AM
Fibonacci
Intro Calc HW
logs----

if $10000 is invested at an interest rate compounded annually the investment will grow to$P after x years where

P=10 000(1.08)^x

d) how long will it take for the ionvestment to grow $50 000 if the interest is 14% for the first 8 yrs and 10% thereafter? e) find the annual interest rate necessary for the$10 000 to double in value in 5 yrs. (give ur answer as percentage)

Thankyu
• Apr 3rd 2008, 04:24 AM
Soroban
Hello, Fibonacci!

Quote:

If $10000 is invested at $r$% compounded annually. the investment will grow to$P after x years where: . $P\:=\:10 000(1+r)^x$

d) How long will it take for the investment to grow $50 000 if the interest is 14% for the first 8 yrs and 10% thereafter? The$10,000 is invested at 14% for the first 8 years.
. . Its value will be: . $10,000(1.14)^8$ dollars.

Then that amount is invested at 10% for $x$ years.
. . The final value is: . $10,000(1.14)^8 \cdot(1.10)^x$
. . which is to equal $50,000. There is our equation . . . $10,000(1.14)^8(1.1)^x \:=\:50,000$ Divide by 10,000: . $(1.14)^8(1.1)^x \:=\:5$ Divide by $(1.14)^8\!:\quad(1.1)^x \:=\:\frac{5}{1.14^8}$ Take logs: . $\ln(1.1)^x \:=\:\ln\left(\frac{5}{1.14^8}\right) \quad\Rightarrow\quad x\!\cdot\!\ln(1.1) \:=\:\ln\left(\frac{5}{1.14^8}\right)$ Therefore: . $x \;=\;\frac{\ln\left(\frac{5}{1.14^8}\right)}{\ln(1 .1)} \;=\;5.88826729 \;\approx\;6$ years. It will take: . $8 + 6 \:=\:\boxed{14\text{ years}}$ Quote: e) Find the annual interest rate necessary for the$10 000
to double in value in 5 yrs. (Give your answer as percentage.)

Let $r$ = the required interest rate.

Then: . $10,000(1 + r)^5 \;=\;20,000$

Divide by 10,000: . $(1+r)^5 \:=\:2$

Take the fifth root: . $1 + r \:=\:\sqrt[5]{2}$

Therefore: . $r \;=\;\sqrt[5]{2}-1 \;=\;0.148698355 \;\approx\;\boxed{14.9\%}$