Exponents

[Explain this!]

How are the following simplified?

1. 5(100 + 0.05)³ = ?

2. (100 + 0.05)³ = ?

3. (100 + 100 * 0.05)³ = ?

 

[Dr.Peterson]

From the inside out.

In each case, evaluate the expression inside the parentheses; then raise that to the power; then do the multiplication on the outside (if any).

If you're unsure, show what you did, and we can check it.

One thing you shouldn't do (though you could if you really wanted to) is to distribute first. Many students, when they see something that resembles [MATH](a + b)^2[/MATH], automatically think they are supposed to distribute (or FOIL, or whatever). But that's only to simplify ("expand") an expression with variables. When you are evaluating with actual numbers, that only gives you more work to do, not less -- especially when the exponent is greater than 2.

 

[Explain this!]

Are these then correct?

1. (500.25)³ = 123187593.765...

2. (100.05)³ = 1001500.750...

3. (105)³ = 1157625

 

[Dr.Peterson]

One of them is wrong, because you didn't do the inside before using anything on the outside.

Which one is it? Can you find the error and fix it?

 

[JeffM]

Are these then correct?

1. (500.25)³ = 123187593.765...

No. You were told to evaluate inside the parentheses first, as indicated by PEMDAS.

[MATH]5(100 + 0.005)^3 = 5 * 100.05^3.[/MATH]
You were then told to exponentiate second as indicated by PEMDAS.

[MATH]5 * 100.05^3 = 5 * 1,001,500.750,125.[/MATH]
And you multiply third as indicated by PEMDAS.

[MATH]5 * 1,001,500.750,125. = 5,007,503.750,625.[/MATH]

2. (100.05)³ = 1001500.750...

3. (105)³ = 1157625

Yes to both

 

[Explain this!]

One of them is wrong, because you didn't do the inside before using anything on the outside.

Which one is it? Can you find the error and fix it?

1. is incorrect. It should be

5 * 1,001,500.750... = 5,007.503.75...

 

[Explain this!]

I have a follow-up question:

If (100 + 100 * 0.05)³ can be simplified to 100(1 + 0.05)³, why is the decimal for the solution for (100 + 100 * 0.05)³ in a different place?

(100 + 100 * 0.05)³ = 1157625, but 100(1 + 0.05)³ = 115.7625

 

[JeffM]

I have a follow-up question:

If (100 + 100 * 0.05)³ can be simplified to 100(1 + 0.05)³, why is the decimal for the solution for (100 + 100 * 0.05)³ in a different place?

(100 + 100 * 0.05)³ = 1157625, but 100(1 + 0.05)³ = 115.7625

[MATH]](100 + 100 * 0.05)^3 = \{100(1 + 0.05)\}^3 =[/MATH]
[MATH]100^3 * 1.05^3 = 1,000,000 * 1.05^3 = 1,157,625.[/MATH]
In other words, everything inside the parentheses is being raised to the power of 3.

[MATH](100 *1.05)^3 = 100^3 * 1.05^3 \ne 100 * 1.05^3.[/MATH]
Do you follow?

 

[Dr.Peterson]

I have a follow-up question:

If (100 + 100 * 0.05)³ can be simplified to 100(1 + 0.05)³, why is the decimal for the solution for (100 + 100 * 0.05)³ in a different place?

(100 + 100 * 0.05)³ = 1157625, but 100(1 + 0.05)³ = 115.7625

The answer is, it can't be simplified that way!

Rather, you could simplify it this way:

(100 + 100 * 0.05)³ = [100(1 + 0.05)]³ = 100³ (1 + 0.05)³ = 1000000 (1 + 0.05)³​

You can't move the 100 outside the parentheses as you did, because that also removed it from the cubing.

This is one reason I recommend always evaluating an expression as given to you, without simplifying: It is too easy to make a mistake in simplifying.

 

[Explain this!]

Why can't (100 + 100* 0.05)³ be simplified as 100(1 + 0.05)³?

The compound interest formula is (P + Pr)ⁿ. This simplifies to P(1 + r)ⁿ. What is the difference between (100 + 100*0.05)³ and (P + Pr)ⁿ = P(1 + r)ⁿ if P is 100?

 

[Dr.Peterson]

Why can't (100 + 100* 0.05)³ be simplified as 100(1 + 0.05)³?

The compound interest formula is (P + Pr)ⁿ. This simplifies to P(1 + r)ⁿ. What is the difference between (100 + 100*0.05)³ and (P + Pr)ⁿ = P(1 + r)ⁿ if P is 100?

You've been told twice why you can't do that. Please read carefully and think about it. When you distribute, you must only take a factor outside of parentheses themselves: (ax + ay) = a(x + y). You can't take a factor outside of a power; it is not true that (ax)ⁿ = a(xⁿ).

So it is not true that (P + Pr)ⁿ simplifies to P(1 + r)ⁿ.

Did you try checking the claim with simple numbers? (I thought you did before.) If P = 100, n = 3, and r = 0.05, then

(P + Pr)ⁿ = (100 + 100*0.05)³ = (105)³ = 1,157,625​

P(1 + r)ⁿ = 100(1 + 0.05)³ = 100(1.05)³ = 100*1.157625 = 115.7625​

Those are different. In the first, 100 is inside the parentheses, so it is cubed; in the second, 100 is outside the parentheses and is not cubed.

 

[Explain this!]

You've been told twice why you can't do that. Please read carefully and think about it. When you distribute, you must only take a factor outside of parentheses themselves: (ax + ay) = a(x + y). You can't take a factor outside of a power; it is not true that (ax)ⁿ = a(xⁿ).

So it is not true that (P + Pr)ⁿ simplifies to P(1 + r)ⁿ.

Did you try checking the claim with simple numbers? (I thought you did before.) If P = 100, n = 3, and r = 0.05, then

(P + Pr)ⁿ = (100 + 100*0.05)³ = (105)³ = 1,157,625​

P(1 + r)ⁿ = 100(1 + 0.05)³ = 100(1.05)³ = 100*1.157625 = 115.7625​

Those are different. In the first, 100 is inside the parentheses, so it is cubed; in the second, 100 is outside the parentheses and is not cubed.

Then how does the compound interest formula become P(1 + r)ⁿ? If I want to determine the compound interest for (100 + 100 * 0.05)³, you are indicating that the 100( 1 + 0.05)³ is incorrect. I understand what you are indicating: "You can't take a factor outside of a power; it is not true that (ax)n = a(x)ⁿ." I do not understand how the compound interest formula became P(1 + r)ⁿ. I thought that I was from (P + Pr)ⁿ, but you indicate that this in not true.

 

[JeffM]

Why can't (100 + 100* 0.05)³ be simplified as 100(1 + 0.05)³?

The compound interest formula is (P + Pr)ⁿ. This simplifies to P(1 + r)ⁿ. What is the difference between (100 + 100*0.05)³ and (P + Pr)ⁿ = P(1 + r)ⁿ if P is 100?

Who in the world told you that the compound interest formula is

[MATH](P + Pr)^n.[/MATH]
That is incorrect unless r = 0 or, as was explained to you in a previous thread, n = 1. The correct formulas are

[MATH]n = 1 \implies E_1 = P_1 + P_1r = P_1(1 + r) = P_1(1 + r)^1.[/MATH]
[MATH]n = 2 \implies E_2 = E_1 + E_1r = E_1(1 + r) = P_1(1 + r)(1 + r) = P_1(1 + r)^2.[/MATH]
[MATH]n = 3 \implies E_3 = E_2 + E_2r = E_2(1 + r) = P_1(1 + r)^2(1 + r) = P_1(1 + r)^3.[/MATH]
In general

[MATH]n > 1 \text { and } r \ne 0 \implies E_n = P_1(1 + r)^n \ne (P_1 + P_1r)^n.[/MATH]

 

[Dr.Peterson]

Then how does the compound interest formula become P(1 + r)ⁿ? If I want to determine the compound interest for (100 + 100 * 0.05)³, you are indicating that the 100( 1 + 0.05)³ is incorrect. I understand what you are indicating: "You can't take a factor outside of a power; it is not true that (ax)n = a(x)ⁿ." I do not understand how the compound interest formula became P(1 + r)ⁿ. I thought that I was from (P + Pr)^{n[/SUP], but you indicate that this in not true.}

Somehow you have it backward. The correct formula is P(1 + r)ⁿ, not (P + Pr)ⁿ. Why do you assume the latter?

Perhaps you need to go back to your textbook or other source and see if you are misreading something.

If you're basing the formula on your own derivation, think again. The idea is that each compounding period, the amount is multiplied by (1 + r); so only that is what is raised to a power.

 

[Explain this!]

Somehow you have it backward. The correct formula is P(1 + r)ⁿ, not (P + Pr)ⁿ. Why do you assume the latter?

Perhaps you need to go back to your textbook or other source and see if you are misreading something.

If you're basing the formula on your own derivation, think again. The idea is that each compounding period, the amount is multiplied by (1 + r); so only that is what is raised to a power.

So, P(1 + r)ⁿ comes from (P + Pr) because (P + Pr) has no exponent. If this is true, when is the exponent "n" used with P( 1 + r)?

 

[Dr.Peterson]

So, P(1 + r)ⁿ comes from (P + Pr) because (P + Pr) has no exponent. If this is true, when is the exponent "n" used with P( 1 + r)?

I would say that P(1 + r)ⁿ comes from P(1 + r). After one compounding period, P has been multiplied by (1 + r). After a second, the result is again multiplied by (1 + r), with the result being p(1 + r)(1 + r) = P(1 + r)². After a third period, it has been multiplied a third time, so the result is P(1 + r)³.

In general, after compounding n times, the amount is P(1 + r)ⁿ.

You have been told this over and over. Why are you not learning?

 

[Explain this!]

I would say that P(1 + r)ⁿ comes from P(1 + r). After one compounding period, P has been multiplied by (1 + r). After a second, the result is again multiplied by (1 + r), with the result being p(1 + r)(1 + r) = P(1 + r)². After a third period, it has been multiplied a third time, so the result is P(1 + r)³.

In general, after compounding n times, the amount is P(1 + r)ⁿ.

You have been told this over and over. Why are you not learning?

If I understand correctly, P(1 + r) simplifies from (P + Pr). P(1 + r)ⁿ is from P( 1 + r)*(1 + r)*(1 +r)*(1 + r). This would be P(1 + r)ⁿ where "n" would equal 3. It could also be expressed as (P + Pr) + r(P + Pr) + r(P + Pr) + r(P +Pr). I am not sure why the "r" is needed before the (P + Pr) as in r(P +Pr).

"Why are you not learning" I am a senior citizen, and I use mathematics to keep my mind active. I sometime struggle with mathematical concepts. I'm sure that you can understand that!

 

[Explain this!]

...where "n" would equal 4.

 

[Dr.Peterson]

It could also be expressed as (P + Pr) + r(P + Pr) + r(P + Pr) + r(P +Pr). I am not sure why the "r" is needed before the (P + Pr) as in r(P +Pr).

I don't think that way of expressing it is helpful (or correct); I think someone may have used it in passing in a previous thread, without intending it to be helpful. It's probably better to ignore it. Clear (like the rest of what you said) is better, when the goal is learning.

But I suspect what you quoted was really (P + Pr) + r(P + Pr) + r((P + Pr) + r(P +Pr)). That makes sense if you put in the effort to untangle it, though no one would ever use it.

By the way, I've often wished that everyone could use their profiles to tell about their personal context, so we could more easily check and see that this "Junior Member" is really a senior, not a high school kid, rather than trying to remember who is who from whatever they've said in passing. But that would be hard to enforce, and easy to abuse.

 

[Explain this!]

I don't think that way of expressing it is helpful (or correct); I think someone may have used it in passing in a previous thread, without intending it to be helpful. It's probably better to ignore it. Clear (like the rest of what you said) is better, when the goal is learning.

But I suspect what you quoted was really (P + Pr) + r(P + Pr) + r((P + Pr) + r(P +Pr)). That makes sense if you put in the effort to untangle it, though no one would ever use it.

By the way, I've often wished that everyone could use their profiles to tell about their personal context, so we could more easily check and see that this "Junior Member" is really a senior, not a high school kid, rather than trying to remember who is who from whatever they've said in passing. But that would be hard to enforce, and easy to abuse.

I want to thank you for the reply and information. I first saw the (P + Pr) + r(P + Pr) at the following: http://moneychimp.com/articles/finworks/fmfutval.htm

I'm not sure what to do after the first r(P + Pr). Is there another r(P + Pr) after that? This calculation works after the first r(P + Pr), but I could not get it to produce the correct compound amount when I added a third r(P + Pr). In your example: (P + Pr) + r(P + Pr) + r((P + Pr) + r(P +Pr)), how would this be solved? The extra parentheses confuse me. You indicated that it (the calculation) makes sense.

As far as "...profiles to tell about their personal context." I will try to indicate that I am not a student in high school or college and that I am just an older adult that is curious to know about some mathematical topic in my future questions that I may post.