indexing ... how to solve (5th-rt(243) * sqrt(64^3))/(27^(2/3) * 512^(-2/3))
- Thread starter kairee
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Has your class not covered exponents and radicals yet, so you need an explanation of the topic? If so, there are loads of lessons available online, such as in this listing.
Please study at least two lessons from the listing, and then attempt the exercise. If you get stuck, you can then reply with a clear listing of your thoughts and efforts, at which point we can begin working with you. ("Read Before Posting")
Thank you!
Steven G
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This is a help forum where we help students solve their problems as compared to doing it for them. Did you read the posting guidelines?
To compute the fifth root of 243 you want to find a single number that when you multiply it by itself 5 times will give you 243. That single number will be your answer.
To compute 272/3 you ask yourself what number when you multiply it by itself 3 times will give you 27. Then you square that answer (square means to multiply by itself 2 times).
If the power/exponent is negative then move it to the other side of the division line, change the sign of the power and simplify as noted above.
Post back with your work using the above hints.
To compute the fifth root of 243 you want to find a single number that when you multiply it by itself 5 times will give you 243. That single number will be your answer.
To compute 272/3 you ask yourself what number when you multiply it by itself 3 times will give you 27. Then you square that answer (square means to multiply by itself 2 times).
If the power/exponent is negative then move it to the other side of the division line, change the sign of the power and simplify as noted above.
Post back with your work using the above hints.
Harry_the_cat
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First things first. Do you recognise that 64 and 512 can both be written as powers of 2 and that 27 and 243 can be written as powers of 3?
Get to know the powers of 2 up to 2^10 and powers of 3 up to 3^5. These numbers crop up all the time in index questions.
Get to know the powers of 2 up to 2^10 and powers of 3 up to 3^5. These numbers crop up all the time in index questions.
The Highlander
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Hi @kairee,If I have followings questions how do I get this to solve , a step details and explanation please.
Do you know anything at all about working with indices?
An index (indices is the plural) or power is the little number you see up and to the right of numbers in your expression.
If you see xn we say x to the power n but for n = 2 or 3 we have special names, thus...
for x2 we say x squared, for x3 we say x cubed but for x4 we say x to the power 4, for x5 we say x to the power 5 and so on.
Now @Harry_the_cat gave you good advice (above) about learning the powers of 2 & 3 up to 210 & 35 respectively and it would be wise of you to take her advice to heart and memorize those results but I want to make sure that you are thoroughly familiar with some even more basic stuff!
Opposite I have included a table of the first 25 squares. It's just a picture (that I grabbed off the internet and modified for my purpose here) so you can right-click on it and Save it for future use. (Please do so.)
Below, however, is a table I have created for you that has a similar construction but this is not a picture so I want you to copy it out and complete it. (That shouldn't be too difficult for you if you study the squares table and see how it has been created.)
Please do that now and post a picture of your work so that we know you have understood everything so far (and correctly completed the table) because only when you have done that will you be in a better position to work on your initial problem.
Regarding the squares table (opposite) I tell my pupils that they should know the first 25 squares and insist that they must learn the first 15 (plus 20 & 25) off by heart. These things are easily found using a calculator but knowing them allows you to spot shortcuts and solutions in Maths problems because the authors of the problems often use these results when writing their questions so that they can be simplified by the students attempting to answer them.
Regarding the table below, I usually just tell my pupils to complete it up to 5 but I have asked you to do so up to 8 for a good reason. In fact, it would do no harm for you to extend the table not only to complete it up to 10 but also by adding a further 4 columns to it: 4th Power (x4); 4th Roots; 5th Power (x5) and 5th Roots. (Doing so would make your initial problem a good deal easier for you to solve!)
No. (x) | Squared (x2) | Square Roots | Cubed (x3) | Cube Roots |
1 | 1 | 1 × 1 | 1 | 1 × 1 × 1 |
2 | 4 | 2 × 2 | 8 | 2 × 2 × 2 |
3 | 9 | 3 × 3 | | |
4 | | | | |
5 | | | | |
6 | | | | |
7 | | | | |
8 |
That completes the 'preliminary' work that you should do before attempting to solve your original expression.
I have illustrated how to proceed from here in the next post (this has to be done in a separate post because there are size limits to individual posts).
You may either complete the work asked of you in this post (and post a picture of your completed table) before reading the next post or you can, if you wish, read on about how to use what you have (I trust?) learned from this post and then post all your work (the table above plus your solution to the original problem) in a single post after you have studied what I write in my next post.
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The Highlander
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@kairee (Continued...)
OK, If you have now come back and posted your work so far, then this is how to proceed from that point forward (with all your new knowledge!).
You may, of course, just be reading on and intend to post your solution to the initial problem at the same time as your completed table (from the previous post).
(You should post a picture of your (handwritten) solution along with your completed table from above.)
So the first thing to know is that when a term has a power (index) that is negative then that means you are dividing by that term so, in effect, a term with a negative index becomes a fraction, eg:-
Now, I trust you know that dividing by a fraction is equivalent to multiplying by its inverse, eg:
\(\displaystyle 2\div\frac{1}{4}=2\times\frac{4}{1}= 2\times 4=8\)
That means that the first simplification you should do to your initial expression is to 'get rid' of that term with the negative index in the denominator by moving it to the numerator, thus...
Now, once you've done that little 'trick' you now have four terms (with powers) that can all be further simplified to make evaluation of the initial expression much easier.
It could easily be done on a calculator, of course, but I suspect that the whole object of this exercise is for you to show that you know how to simplify these terms to make the final calculation much more straightforward.
This is where the work you have done above (assuming you have done it, of course) comes into play.
I will explain how to do one of them for you (with just hints for the other three).
You know (from the earlier work I set you) just how squares (& cubes) 'work', so let's look at: \(\displaystyle \sqrt{64^3}\)
\(\displaystyle \qquad\sqrt{64^3}=\sqrt{64\times 64\times 64}\\\,\\=\sqrt{(8\times 8)\times (8\times 8)\times (8\times 8)}\\\,\\=\sqrt{(8\times 8)\times 8\times 8\times (8\times 8)}\\\,\\=\sqrt{((8\times 8)\times 8)\times (8\times (8\times 8))}\\\,\\=\sqrt{(64\times 8)\times (8\times 64)}\\\,\\=(64\times 8)=\text{What?}\)
Do you see what I did there?
You know (from your earlier efforts) that 643 is 64 × 64 × 64 but you also know that 64 is a square number (8 × 8) so I replaced each 64 (under the square root sign) with (8 × 8), then I opened up the middle (8 × 8) and associated its 8's with the first & last (8 × 8) which allowed me to have just two (the same) terms multiplying each other under the square root sign, so I can simply remove the radical (that's the square root sign) because that term is the square root I'm looking for.
Here is a (similar) fully worked example...
\(\displaystyle \qquad\sqrt{16^3}=\sqrt{16\times 16\times 16}\\\,\\=\sqrt{(4\times 4)\times (4\times 4)\times (4\times 4)}\\\,\\=\sqrt{(4\times 4)\times 4\times 4\times (4\times 4)}\\\,\\=\sqrt{((4\times 4)\times 4)\times (4\times (4\times 4))}\\\,\\=\sqrt{(16\times 4)\times (4\times 16)}\\\,\\=(16\times 4)=64\)
Right that's one technique you need to use but what about the terms with the fractional indices (powers)?
Well the only fractional power you have left to deal with now is ⅔; so what does that mean?
\(\displaystyle x^{\frac{2}{3}}\) means that you need to find the square of the cube root of x or the cube root of x squared.
It doesn't matter whether you square x first and then find the cube root of that result or you find the cube root of x and then square that result; you will arrive at the same answer either way.
So, if you know the cube root of 27 (hint), then you only have to square that to get \(\displaystyle 27^{\frac{2}{3}}\) (and the same applies to \(\displaystyle 512^{\frac{2}{3}}\)).
Do you see now why it was so important for you to do all that work on the 'tables' above?
That just leaves you with \(\displaystyle \sqrt[5]{243}\) to deal with.
Well \(\displaystyle \sqrt[5]{x}\) (as I hope you have already figured out) just means that \(\displaystyle \sqrt[5]{x}=\sqrt[5]{a\times a\times a\times a\times a}\text{ where } x=a^5\)
So how can you factorize (split up) the number under the radical (243) into a multiplication sequence like you have done above (where all five numbers in the sequence are the same)?
Do you know the 'trick' for determining whether a number is a multiple of 9?
If the digits in a number add up to nine then the number can be divided evenly by 9, eg;
333 add to 9 so 333 is a multiple of 9 (333 ÷ 9 = 37 and 9 × 37 = 333)
Similarly, 666 add to 18 which add to 9 so 666 is a multiple of 9 (666 ÷ 9 = 74 and 9 × 74 = 666).
So what do the digits of 243 add to (and how does that help you)? Think about it.
Please now come back and confirm you have done the work I set you as preliminary work by showing us your completed table of cubes (and fourth & fifth powers if you extended it, as suggested).
Followed up separately (or together) by your simplification of the initial expression to numbers that have no powers involved so that they can be easily multiplied and divided to get the final result for the evaluation of the original expression (which should be: 10,922⅔ ≈ 10,922.67).
You should show the steps (like I have) that you go through to simplify the original terms (to 'eliminate' the indices) so there will be at least three or four lines in your working.
Please write your working out neatly and legibly for us to check and a picture of your work uploaded to the site will suffice for that.
We look forward to hearing from you soon.
Hope that helps.
OK, If you have now come back and posted your work so far, then this is how to proceed from that point forward (with all your new knowledge!).
You may, of course, just be reading on and intend to post your solution to the initial problem at the same time as your completed table (from the previous post).
(You should post a picture of your (handwritten) solution along with your completed table from above.)
So the first thing to know is that when a term has a power (index) that is negative then that means you are dividing by that term so, in effect, a term with a negative index becomes a fraction, eg:-
\(\displaystyle x^{-2}=\frac{1}{x^2};\qquad x^{-3}=\frac{1}{x^3}\qquad\text{and}\qquad 512^{-\frac{2}{3}}=\frac{1}{512^{\frac{2}{3}}}\)
(As @Steven G tried to explain in his post above, qv.)
(As @Steven G tried to explain in his post above, qv.)
Now, I trust you know that dividing by a fraction is equivalent to multiplying by its inverse, eg:
\(\displaystyle 2\div\frac{1}{4}=2\times\frac{4}{1}= 2\times 4=8\)
That means that the first simplification you should do to your initial expression is to 'get rid' of that term with the negative index in the denominator by moving it to the numerator, thus...
\(\displaystyle \frac{\sqrt[5]{243}\times\sqrt{64^3}}{27^{\frac{2}{3}}\times 512^{-\frac{2}{3}}}=\frac{\sqrt[5]{243}\times\sqrt{64^3}}{27^{\frac{2}{3}}\times\frac{1}{512^{\frac{2}{3}}}}=\frac{\sqrt[5]{243}\times\sqrt{64^3}\times 512^{\frac{2}{3}}}{27^{\frac{2}{3}}}\)
Now, once you've done that little 'trick' you now have four terms (with powers) that can all be further simplified to make evaluation of the initial expression much easier.
It could easily be done on a calculator, of course, but I suspect that the whole object of this exercise is for you to show that you know how to simplify these terms to make the final calculation much more straightforward.
This is where the work you have done above (assuming you have done it, of course) comes into play.
I will explain how to do one of them for you (with just hints for the other three).
You know (from the earlier work I set you) just how squares (& cubes) 'work', so let's look at: \(\displaystyle \sqrt{64^3}\)
\(\displaystyle \qquad\sqrt{64^3}=\sqrt{64\times 64\times 64}\\\,\\=\sqrt{(8\times 8)\times (8\times 8)\times (8\times 8)}\\\,\\=\sqrt{(8\times 8)\times 8\times 8\times (8\times 8)}\\\,\\=\sqrt{((8\times 8)\times 8)\times (8\times (8\times 8))}\\\,\\=\sqrt{(64\times 8)\times (8\times 64)}\\\,\\=(64\times 8)=\text{What?}\)
Do you see what I did there?
You know (from your earlier efforts) that 643 is 64 × 64 × 64 but you also know that 64 is a square number (8 × 8) so I replaced each 64 (under the square root sign) with (8 × 8), then I opened up the middle (8 × 8) and associated its 8's with the first & last (8 × 8) which allowed me to have just two (the same) terms multiplying each other under the square root sign, so I can simply remove the radical (that's the square root sign) because that term is the square root I'm looking for.
Here is a (similar) fully worked example...
\(\displaystyle \qquad\sqrt{16^3}=\sqrt{16\times 16\times 16}\\\,\\=\sqrt{(4\times 4)\times (4\times 4)\times (4\times 4)}\\\,\\=\sqrt{(4\times 4)\times 4\times 4\times (4\times 4)}\\\,\\=\sqrt{((4\times 4)\times 4)\times (4\times (4\times 4))}\\\,\\=\sqrt{(16\times 4)\times (4\times 16)}\\\,\\=(16\times 4)=64\)
Right that's one technique you need to use but what about the terms with the fractional indices (powers)?
Well the only fractional power you have left to deal with now is ⅔; so what does that mean?
\(\displaystyle x^{\frac{2}{3}}\) means that you need to find the square of the cube root of x or the cube root of x squared.
It doesn't matter whether you square x first and then find the cube root of that result or you find the cube root of x and then square that result; you will arrive at the same answer either way.
So, if you know the cube root of 27 (hint), then you only have to square that to get \(\displaystyle 27^{\frac{2}{3}}\) (and the same applies to \(\displaystyle 512^{\frac{2}{3}}\)).
Do you see now why it was so important for you to do all that work on the 'tables' above?
That just leaves you with \(\displaystyle \sqrt[5]{243}\) to deal with.
Well \(\displaystyle \sqrt[5]{x}\) (as I hope you have already figured out) just means that \(\displaystyle \sqrt[5]{x}=\sqrt[5]{a\times a\times a\times a\times a}\text{ where } x=a^5\)
So how can you factorize (split up) the number under the radical (243) into a multiplication sequence like you have done above (where all five numbers in the sequence are the same)?
Do you know the 'trick' for determining whether a number is a multiple of 9?
If the digits in a number add up to nine then the number can be divided evenly by 9, eg;
333 add to 9 so 333 is a multiple of 9 (333 ÷ 9 = 37 and 9 × 37 = 333)
Similarly, 666 add to 18 which add to 9 so 666 is a multiple of 9 (666 ÷ 9 = 74 and 9 × 74 = 666).
So what do the digits of 243 add to (and how does that help you)? Think about it.
Please now come back and confirm you have done the work I set you as preliminary work by showing us your completed table of cubes (and fourth & fifth powers if you extended it, as suggested).
Followed up separately (or together) by your simplification of the initial expression to numbers that have no powers involved so that they can be easily multiplied and divided to get the final result for the evaluation of the original expression (which should be: 10,922⅔ ≈ 10,922.67).
You should show the steps (like I have) that you go through to simplify the original terms (to 'eliminate' the indices) so there will be at least three or four lines in your working.
Please write your working out neatly and legibly for us to check and a picture of your work uploaded to the site will suffice for that.
We look forward to hearing from you soon.
Hope that helps.
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