Indices Question

jonnburton

Junior Member
Joined
Dec 16, 2012
Messages
155
Hi,

I've been working on indices and am having a little difficulty fully understanding it. The following is an example of a question which I got wrong and am unable to see why. I would be grateful for any indicators as to where my thinking is off track.

Simplify the following:

\(\displaystyle 16a^-2b^2c^-3 * 2 (abc)^-2\)

Multiplying out the brackets gives:

16a^-2 * b^2 * c^-3 * 2a^-2 *2b^-2 * 2c^-2

= 32a^-4 * 2b^0 * 2c^-5

However, the answer the book gives is:

32a^-4b^0c^-5

My question here is, what happens to the 2 before the brackets in the original equation; why does this dissapear? (I would have thought that b^2 * 2b^-2 = 2b^0, and c^-3 * 2c^-2 = 2c^-5 ...)
 
Hi,

I've been working on indices and am having a little difficulty fully understanding it. The following is an example of a question which I got wrong and am unable to see why. I would be grateful for any indicators as to where my thinking is off track.

Simplify the following:

\(\displaystyle 16a^-2b^2c^-3 * 2 (abc)^-2\)

Multiplying out the brackets gives:

16a^-2 * b^2 * c^-3 * 2a^-2 *2b^-2 * 2c^-2

= 32a^-4 * 2b^0 * 2c^-5

However, the answer the book gives is:

32a^-4b^0c^-5

My question here is, what happens to the 2 before the brackets in the original equation; why does this dissapear? (I would have thought that b^2 * 2b^-2 = 2b^0, and c^-3 * 2c^-2 = 2c^-5 ...) Your statement is in the conditional. Under what circumstances do you find 16 * 2 = 32 difficult to understand?
Is, by an remote chance, the problem to simplify

\(\displaystyle 16a^{-2} * b^2 * c^{-3} * 2(abc)^{-2}.\)

The book's answer looks fine though incomplete to me.

\(\displaystyle 16a^{-2} * b^2 * c^{-3} * 2(abc)^{-2} = (16 * 2)(a^{-2}b^2c^{-3}) * (a^{-2}b^{-2}c^{-2}) = 32a^{(- 2 - 2)}b^{(2 - 2)} * c^{(- 3 - 2)} = 32a^{-4}b^0c^{-5} = \dfrac{32}{a^4c^5}.\)
 
Last edited:
Thanks for your reply, Jeff.

Sorry, I didn't check that the original question was formatted correctly in latex. (having difficulty formatting the \(\displaystyle ^{-2}\) was why I stopped doing it in latex).

But yes, that was the original problem.

My problem wasn't in understanding that 16*2 =32, it was more the order to do the operations in the equation. The way I would have tried to solve this (before having seen how you arrived at your solution) would have been to do the following:

\(\displaystyle 16a^{-2}*b^2*c^{-3} * 2(abc)^{-2}\)

\(\displaystyle 16a^{-2}*b^2*c^{-3} * 2a^{-2}*2b^{-2}*2c^{-2}\)

\(\displaystyle 32a^{-4}*2b^0*2c^{-5}\)


It seems to me that the order in which things are done here is different to the way I normally deal with equations. Normally I would start by multiplying out the brackets first - i.e. I would expand the \(\displaystyle 2(abc)^{-2}\) to give \(\displaystyle 2a^{-2}*2b^{-2}*2c^{-2}\)

I see that the correct procedure is to multiply the 2 by the 16, despite the fact that the 2 is placed in front of (and must refer to) the \(\displaystyle (abc)^{-2}\)

I know I am missing something elementary here, but I don't know what it is!
 
Thanks for your reply, Jeff.

Sorry, I didn't check that the original question was formatted correctly in latex. (having difficulty formatting the \(\displaystyle ^{-2}\) was why I stopped doing it in latex).

But yes, that was the original problem.

My problem wasn't in understanding that 16*2 =32, it was more the order to do the operations in the equation. The way I would have tried to solve this (before having seen how you arrived at your solution) would have been to do the following:

\(\displaystyle 16a^{-2}*b^2*c^{-3} * 2(abc)^{-2}\)

\(\displaystyle 16a^{-2}*b^2*c^{-3} * 2a^{-2}*2b^{-2}*2c^{-2}\)

\(\displaystyle 32a^{-4}*2b^0*2c^{-5}\)


It seems to me that the order in which things are done here is different to the way I normally deal with equations. Normally I would start by multiplying out the brackets first - i.e. I would expand the \(\displaystyle 2(abc)^{-2}\) to give \(\displaystyle 2a^{-2}*2b^{-2}*2c^{-2}\)

I see that the correct procedure is to multiply the 2 by the 16, despite the fact that the 2 is placed in front of (and must refer to) the \(\displaystyle (abc)^{-2}\)

I know I am missing something elementary here, but I don't know what it is!
In LaTeX, you can superscribe an exponent expressible in ONE character by preceding that character by ^. So you can write x2 as x^2. But if you want to superscribe an exponent that requires more than one character to express, you must follow the ^ with {expression}. To show
x(-10), you must write x^{(-10)}. Same rules apply to subscripts.

Adding exponents does not change the basic rules of algebra that you have already learned. It is a psychological quirk of the human mind that when you are concentrating on learning something new, something long known sometimes slips your mind. It is an embarrassment that happened to me just this weekend.

(abc) is just a number that has three factors.

\(\displaystyle abc = d \ne 0 \implies 2(abc) = 2d \ne 8d = 2a * 2b * 2c.\)

Let's try an experiment.

\(\displaystyle a = 3,\ b = 5\ ,\ and\ c = 7 \implies abc = 105.\)

\(\displaystyle So\ 2(abc) = 2 * 105 = 210.\)

\(\displaystyle But\ 2a = 6,\ 2b = 10, 2c = 14 \implies 2a * 2b * 2c = 840 \ne 210.\)

The rules of algebra are just the rules of arithmetic made compleltely general.
 
Thank you very much for explaining that Jeff. I can see what the problem was very clearly now - and it makes perfect sense!

It is amazing how our minds can temporarily forget such basic things but that's one mistake I won't be making again!
 
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