# Square root and substitution

1. Mar 4, 2008

### iksotof

Have this question in relation to some investment exam I am doing, I am a maths novice being some years since leaving school etc, ok enough of the excuses.

Example

FV = future value
PV = present value
R = interest rate
N = number of compounding periods

my PV is 6000 and my FV is 10000. Compounding periods is 12, I need to find the interest rate, thus...

100000 = 6000 (1 + R) to power of 12 (sorry dont know how to represent that on key board).

substitution: 1.67 = (1 + R) power 12

substitution 2: 12√1.67 = 1 + R

1.0435 - 1 = R

R = 4.45.

All very well, but I lose understanding of how the maths is applied at sub 2, my text book
has given this model answer. For a novice please show me step by step what

12√1.67 means.

Thanks you Darren.

2. Mar 4, 2008

### arildno

Okay, to "cancel" the operation of cubing a number, you need to to take the cube root in order to get back to your number.

That is we have for any number a:
$$\sqrt[3]{a^{3}}=a$$
(Taking the cube root of a cube leaves you with the original number "a")

Similarly, if a positive number "a" is raised to the twelfth power, you must take the twelfth root to get a back:
$$\sqrt[12]{a^{12}}=a.$$

Now, how to use this in your present case`

WEll, you are to solve for R, but that appears in your equation within the expression $(1+R)^{12}[/tex] Now, you are ALLOWED with an equation to "do" whatever operation you like, AS LONG AS YOU DO THE SAME ON BOTH SIDES!. Thus, in order to remove the twelfth power on the right hand side (increasing your chances to solve for R!), you take the twelfth root of both sides. That's what the second substitution does. 3. Mar 4, 2008 ### HallsofIvy Staff Emeritus Using just the keyboard, 100000= 6000(1+R)^(12). Using "html" tags, 100000= 6000(1+R)12. Using "LaTex" [itex]100000= 6000(1+ R)^{12}$. But you don't mean "100000" on the left, only 10000.

Dividing both sides by 6000, 10000/6000= 10/6= 4/3 which is 1.67 to two decimal places.

12√1.67 means.[/quote]
It means the "twelfth root of 1.67", not "square root"! The "twelfth root of x" is defined as the number whose 12 power is x- that's why $^{12}\sqrt{(1+R)^{12}}= !+R$. To actually do the that on a calculator, if you have a "^", power, key, use the "1/12 power": 1.67 ^ (1/12). If your calculator does not have a "^" key you will need to use logarithms. log(x^{1/12})= (1/12)log x= log x/12. To find the 12th root of 1.67, take the logarithm of 1.67, divide by 12, the use the "inverse" to logarithm: 10^ if you are using common logs

4. Mar 4, 2008

### arildno

$^{12}\sqrt{(1+R)^{12}}$, rather: $\sqrt[12]{(1+R)^{12}}$

5. Mar 4, 2008

### iksotof

Thank you greatly but I am a total novice and whilst I am sure your explanation is clear to a knowledgeable person, I am not, in isolation therfore it doesn't mean a great deal. Could you maybe show the maths in long hand please? For example 2 cubed is 2 x 2 x 2=
8.

Please continue with this long hand through the remainder.

Thanks again Darren.

6. Mar 4, 2008

### arildno

Yes, and tell me:

What is the definition of the cube root of a number?

7. Mar 4, 2008

### HallsofIvy

Staff Emeritus
Continue with what? 23= 8 exactly as you said. If you want us to now find $\sqrt[3]{8}$ (thanks, arildno)? There is no calculation involved. By the definition of cube root, it is 2.

If you were to ask me, "What is the cube root of 7?", I would whip out my calculator and answer that $\sqrt[3]{7}= 1.913$, approximately.

8. Mar 4, 2008

### iksotof

Thank you, it was the fundemental understanding I was being a total dumb ass with. . Thanks for your patience and inputs, my old school lessons are finally reawkening from the recesses of my mind, 20 years on!

Greatly appreciated, Darren.

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