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Thread: Intro Calc HW

  1. #1
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    Intro Calc HW

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    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
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  2. #2
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    Hello, Fibonacci!

    If $10000 is invested at $\displaystyle r$% compounded annually.
    the investment will grow to $P after x years where: .$\displaystyle 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 $10,000 is invested at 14% for the first 8 years.
    . . Its value will be: .$\displaystyle 10,000(1.14)^8$ dollars.

    Then that amount is invested at 10% for $\displaystyle x$ years.
    . . The final value is: .$\displaystyle 10,000(1.14)^8 \cdot(1.10)^x$
    . . which is to equal $50,000.

    There is our equation . . . $\displaystyle 10,000(1.14)^8(1.1)^x \:=\:50,000$

    Divide by 10,000: .$\displaystyle (1.14)^8(1.1)^x \:=\:5$

    Divide by $\displaystyle (1.14)^8\!:\quad(1.1)^x \:=\:\frac{5}{1.14^8} $

    Take logs: . $\displaystyle \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: . $\displaystyle x \;=\;\frac{\ln\left(\frac{5}{1.14^8}\right)}{\ln(1 .1)} \;=\;5.88826729 \;\approx\;6$ years.

    It will take: .$\displaystyle 8 + 6 \:=\:\boxed{14\text{ years}}$




    e) Find the annual interest rate necessary for the $10 000
    to double in value in 5 yrs. (Give your answer as percentage.)
    Let $\displaystyle r$ = the required interest rate.

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

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

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

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

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