SOS: simplifying, evaluating, compound interest, etc.

Rohan

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Nov 29, 2006
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Is there anyine out there that could help me with and explain to me the following Questions to get me going.

Question 1: Simplify 125^1/3*5^2/5^3*25^2/5

step 1 ) \sqrt{125}*5^2/5^3*\sqrt{5}
step 2 ) 5*25/5^3(25^1/2)^5
step 3 ) 5(25)/5^3(5)^5
step 4 ) 125/125(125)
= 1/125

the answer I get is =1/125

Question 2: Evaluate (1.5*10^3)x(6*10^10)x(2*10^-4)/(4.5*10^3)x(2*10^3) Give the answer in the form a*10^b where 1.00<a<9.99

I do not know how to start or do this ?

Question 3: If X=(0.98)^3 and Y=(1.01)^4 use the binomial theorem to calculate Z=X+Y to three decimal places.

Question 4: A person has a sum of €6000 to invest.

i) If he invests the money at a rate of 4% compound interest p/a calculate the value of the investment at the end of the fouth year ?

This is the only answer I could get P=(1+R)^n
P=6000
R=0.04
n=4 and the answer is = 7019.15
The rest I don't get i know it would involve a log, but what is where ? LogxY

ii) As an alternative he has been offered an investment whereby the money will amount to 7500 after 5 years. Calc the effective cpmpound interest rate in this inveestment ?

P=7500
R=?
n=5
7500=(1+R)^5
7500=5+5x
7500/5=5+x
1500-5=x
1495=x (somehow I don't think this is the correct answer) HELP :roll:

iii) How long would the original investment, at a rate of 4% (0.04) compound interest p/a have to be, invested for it to double in value ?

P=12000
R=0.04
n=?
12000=(1+0.04)^? (don't know see the bank manager) :D

Cheers for any help and moral support !! I need it !!

I have more questions about matrices ... just need to get this out the way first ... if you want you can also chat to me via MSN ....

Rohan
 

stapel

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Rohan said:
Question 1: Simplify 125^1/3*5^2/5^3*25^2/5
I'm sorry, but I can't figure out what you mean by this. Please reply using grouping symbols to clarify which bits are exponents, which bits are numerators, which bits are multiplied together, etc.

Rohan said:
Question 2: Evaluate (1.5*10^3)x(6*10^10)x(2*10^-4)/(4.5*10^3)x(2*10^3)
Give the answer in the form a*10^b where 1.00<a<9.99

I do not know how to start or do this ?
The "Give the answer in the form" part just means to use scientific notation in your answer. The rest is simple multiplication and application of exponent rules.

I can't be more specific, though, until grouping symbols specify which bits are in the numerator, and which are in the denominator. Is everything before the "slash" in the numerator, and all the rest in the denominator, or is there a fractional term in the middle of the product?

Rohan said:
Question 3: If X=(0.98)^3 and Y=(1.01)^4 use the binomial theorem to calculate Z=X+Y to three decimal places.
Use the Binomial Theorem twice, expanding (1 - 0.2)<sup>3</sup> and (1 + 0.1)<sup>4</sup>.

Rohan said:
Question 4: A person has a sum of €6000 to invest.

i) If he invests the money at a rate of 4% compound interest p/a calculate the value of the investment at the end of the fouth year ?
Since "n" stands for the number of compoundings per year, and since this exercise stipulates "per annum" (once a year), n = 1. But there are four years, so t = 4.

. . . . .A = P(1 + r/n)<sup>nt</sup>

You are looking for the ending amount "A", not the principle (beginning amount) "P". Just plug-n-chug in the compound-interest formula (which you should memorize, if you haven't already), with P = 6000, n = 1, t = 4, and r = 0.04.

Rohan said:
ii) As an alternative he has been offered an investment whereby the money will amount to 7500 after 5 years. Calc the effective cpmpound interest rate in this inveestment ?
Use the same compounding formula, with A = 7500, P = 6000, n = 1, and t = 5. Solve for the interest rate r.

Rohan said:
iii) How long would the original investment, at a rate of 4% (0.04) compound interest p/a have to be, invested for it to double in value ?
Use P = 6000, A = 12000, r = 0.04, and n = 1. Solve for the time t.

Eliz.
 

Rohan

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Joined
Nov 29, 2006
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Thanx :D M8

I shall start working on it ... not all is clear, but enough to get a start.

I have plenty more questions, but they will come in time, your formula makes sence,
Thank you for your quick responce.

G'Day M8 :D
 

soroban

Elite Member
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Jan 28, 2005
Messages
5,588
Hello, Rohan!

Here's some help . . .

1) Simplify: \(\displaystyle 125^{\frac{1}{3}}\,\cdot \,5^{\frac{2}{5}}\,\cdot \,25^{\frac{2}{5}}\)
We have: \(\displaystyle \left(5^3 \right)^{\frac{1}{3}}\,\cdot \,5^{\frac{2}{5}}\,\cdot \,\left(5^2\right)^{\frac{2}{5}} \;=\;5^1\,\cdot \,5^{\frac{2}{5}}\,\cdot \,5^{\frac{4}{5}} \;=\;5^{(1+\frac{2}{5}+\frac{4}{5})} \;=\;5^{\frac{11}{5}}\)

2) Evaluate: \(\displaystyle \frac{ \left(1.5\,\cdot 10^3\right)\,\left(6\,\cdot \,10^{10}\right)\,\left(2\,\cdot 10^{-4}\right) }{ \left(4.5\,\cdot 10^3\right)\,\left(2\,\cdot 10^3\right) }\)

Give the answer in the form \(\displaystyle a\,\cdot\,10^b\) where \(\displaystyle 1\,\leq\,a\,<\,10\)
It's mainly arithmetic and some Algebra I . . .

We have: \(\displaystyle \,\frac{(1.5)(6)(2)}{(4.5)(2)}\,\cdot\,\frac{10^3\,\cdot\,10^{10}\,\cdot\,10^{-4}}{10^3\,\cdot\,10^3} \;=\;\frac{18}{9}\,\cdot\,\frac{10^9}{10^3} \;=\;2\,\cdot\,10^6\)


3) If \(\displaystyle X\,=\,( 0.98 )^3\) and \(\displaystyle Y\,=\,(1.01)^4\)
use the Binomial theorem to calculate \(\displaystyle Z\,=\,X\,+\,Y\) to three decimal places.
We're supposed to notice that those two numbers are very close to 1.

So: \(\displaystyle \,X \;=\;(1\,-\,0.02)^3\)
. . . . . . \(\displaystyle = \;1^3\,-\,3\cdot1^2(0.2)\,+\,3\cdot1(0.02)^2 \,-\,(0.02)^3\)
. . . . . . \(\displaystyle = \;1\,-\,0.06\,+\,0.0012\,-\,0.000008\)
. . . . . . \(\displaystyle = \;0.941192\)

And: \(\displaystyle \,Y\;=\;(1\,+\,0.01)^4\)
. . . . . . .\(\displaystyle = \;1^4\,+\,4\cdot1^3(0.01)\,+\,6\cdot1^2(0.01)^2\,+\,4\cdot1(0.01)^3\,+\,(0.01)^4\)
. . . . . . .\(\displaystyle =\;1\,+\,0.04\,+\,0.0006\,+\,0.000004\,+\,0.00000001\)
. . . . . . .\(\displaystyle = \;1.4060401\)

Therefore: \(\displaystyle \,Z\:=\:X\,+\,Y\:=\:0.941192\,+\,1.04060401\:=\:1.98179601\;\approx\:\L1.982\)


 
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Denis

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Feb 17, 2004
Messages
1,489
Rohan said:
Question 1: Simplify 125^1/3*5^2/5^3*25^2/5
Better be clearer, Rohan, else nobody can help:

125^(1/3) * (5^2 / 5^3) * 25^(2/5)

Is that it?
 

Rohan

New member
Joined
Nov 29, 2006
Messages
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Input

What is the format for inputing problems to make sence ?
Latex ... tried it didn't work ? how do I express my problems ?

Some Clear some not,

Firstly thank you for taking the time to respond, it is much needed and welcome !

Soroban

just a few things in Question 2 (evaluate ...) between the brackets (...*10^3)x(...*10^10) there are are x's or do they cancel each other out ?

Please explain to me Answer 3 (the logic behind it, I do not understand) ?
x=(1-0.02)^3 line 2

y=(1+0.01)^4 line 2

Rohan
 

Rohan

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Joined
Nov 29, 2006
Messages
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Clarity

Yes, you are correct Dennis,

Let me try to explain :
125^(1/3)*5^2
------------------
5^3*25^5/2

In my simple terms :

[125 to the power of 1/3 * 5 to the 2nd power devide by 5 to the 3rd power * 25 to the power of 5/2]

How do I input this Mathematically ?

R
 

Denis

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Joined
Feb 17, 2004
Messages
1,489
I answered your PM on this...
 
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