do not understand, need it to continue to next chapter..

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bobisaka

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answer:
1627596096893.png


My solution:

Apply Quotient Property rule..

((25m^1/6 - ^4/6)/(n^-1/6 -11/6))^1/2

((25m^-3/6)/(n^12/6 ))^1/2

((25m^-1/2)/(n^2/1 ))^1/2

I believe I am stuck at this point, however I will do what i think...

((25)^1/2 (m^-1/2)^1/2)/(n^2/1)^1/2)

Simplify.. square integer & Apply product to power rule..

((sqr 25) (m^-1/2)^1/2)/((n^2/1)^1/2) <----- still unsure when to use square root vs 25*1/2, which equals 12.5
((5) (m^-2/2)/(n^3/2))

(5m)/(n^3/2)


And there you have, still lost as usual. I need to understand what I am doing wrong, before i can continue to next chapter..

How did 5 combine with the 'n' variable, when 25 was connected to 'm'.. so lost.
 
bobisaka said:
View attachment 28428
answer:
View attachment 28429


My solution:

Apply Quotient Property rule..

((25m^1/6 - ^4/6)/(n^-1/6 -11/6))^1/2

((25m^-3/6)/(n^12/6 ))^1/2

((25m^-1/2)/(n^2/1 ))^1/2

I believe I am stuck at this point, however I will do what i think...

((25)^1/2 (m^-1/2)^1/2)/(n^2/1)^1/2)

Simplify.. square integer & Apply product to power rule..

((sqr 25) (m^-1/2)^1/2)/((n^2/1)^1/2) <----- still unsure when to use square root vs 25*1/2, which equals 12.5
((5) (m^-2/2)/(n^3/2))

(5m)/(n^3/2)


And there you have, still lost as usual. I need to understand what I am doing wrong, before i can continue to next chapter..

How did 5 combine with the 'n' variable, when 25 was connected to 'm'.. so lost.
Click to expand...
Your notation may be absolutely correct but it's impossible (for me) to read your work, sorry. Can you use Latex or write on paper and post a photo?

Regarding "still unsure when to use square root vs 25*1/2, which equals 12.5" - What do you mean? The operation is either a square root (the exponent is 1/2) or multiplication, there is no ambiguity whatsoever.
 
Go this way, just one more transformation and there u will be!
[math]\left(\frac{25m^{\frac{1}{6}}n^{\frac{11}{6}}}{m^{\frac{2}{3}}n^{-\frac{1}{6}}}\right)^\frac{1}2{}=25^\frac{1}{2}\cdot\left(\frac{m^\frac{1}{6}}{m^\frac{2}{3}}\right)^\frac{1}{2}\cdot\left(\frac{n^\frac{11}{6}}{n^{-\frac{1}{6}}}\right)^\frac{1}{2}[/math] ........... corrected
 
Last edited by a moderator:
lev888 said:
Your notation may be absolutely correct but it's impossible (for me) to read your work, sorry. Can you use Latex or write on paper and post a photo?
Click to expand...

Unsure how to use LaTex.. I will look into it. I was wondering how you guys write notations..

lev888 said:
Regarding "still unsure when to use square root vs 25*1/2, which equals 12.5" - What do you mean? The operation is either a square root (the exponent is 1/2) or multiplication, there is no ambiguity whatsoever.
Click to expand...

My question maybe, when should i use SQUARE ROOT? In canvas' example above, the first part of the equation says 25^1/2.
Do I square it?
Or
Do I multiply the integer to the exponent?
Which and why?
 
I like to simplify inside grouping symbols, first.

There is a property of exponents for simplifying a ratio of powers having the same base.

[math]\frac{x^h}{x^k} = x^{h-k}[/math]
For example, here is the ratio of powers of m simplified:

[math]\frac{m^{1/6}}{m^{2/3}} = m^{1/6 - 2/3} = m^{-1/2} = \frac{1}{m^{1/2}}[/math]
?
 
Otis said:
I like to simplify inside grouping symbols.

There is a property of exponents for simplifying a ratio of powers having the same base.

[math]\frac{x^h}{x^k} = x^{h-k}[/math]
[math]\frac{m^{1/6}}{m^{2/3}} = m^{1/6 - 2/3} = m^{-1/2} = \frac{1}{m^{1/2}}[/math]
?
Click to expand...

When applying the quotient property in the first part, are there any circumstances in which i move the numerator to the denominator, so that the denominator minus' the numerator?
Because that is how I have been doing it (followed it according to some examples in the text). In regards to your particular example, when i did the result was a positive so i could not further make it into a quotient.
Is there a particular order to this? Is my understanding incorrect?

Moving on, so with the equation m^{-1/2}, the negative property becomes applies the negative property exponent to become the quotient right?
 
bobisaka said:
In regards to your particular example, when i did the result was a positive
Click to expand...
Please show your work.

bobisaka said:
circumstances in which i move the numerator to the denominator, so that the denominator minus' the numerator
Click to expand...
The property is as I posted it. I don't understand the English above.

bobisaka said:
the negative property becomes applies the negative property exponent to become the quotient right?
Click to expand...
I don't understand what you're trying to say.

:confused:
 
bobisaka said:
Unsure how to use LaTex.. I will look into it. I was wondering how you guys write notations..



My question maybe, when should i use SQUARE ROOT? In canvas' example above, the first part of the equation says 25^1/2.
Do I square it?
Or
Do I multiply the integer to the exponent?
Which and why?
Click to expand...
1/2 is the exponent. It's the same as taking a square root. There is nothing even remotely related to multiplication of 25 by 1/2.
 
mmm4444bot said:
The OP may be confusing [imath]25m^{1/2}[/imath] with [imath](25m)^{1/2}[/imath] and isn't being clear at explaining thoughts.

?
Click to expand...
"still unsure when to use square root vs 25*1/2, which equals 12.5"
 
bobisaka said:
View attachment 28428
answer:
View attachment 28429


My solution:

Apply Quotient Property rule..

((25m^1/6 - ^4/6)/(n^-1/6 -11/6))^1/2

((25m^-3/6)/(n^12/6 ))^1/2

((25m^-1/2)/(n^2/1 ))^1/2

I believe I am stuck at this point, however I will do what i think...

((25)^1/2 (m^-1/2)^1/2)/(n^2/1)^1/2)

Simplify.. square integer & Apply product to power rule..

((sqr 25) (m^-1/2)^1/2)/((n^2/1)^1/2) <----- still unsure when to use square root vs 25*1/2, which equals 12.5
((5) (m^-2/2)/(n^3/2))

(5m)/(n^3/2)


And there you have, still lost as usual. I need to understand what I am doing wrong, before i can continue to next chapter..

How did 5 combine with the 'n' variable, when 25 was connected to 'm'.. so lost.
Click to expand...
Hi Bobisaka. If you don't know how to use LaTex, just take a photo of your handwritten work and post it. It is very hard to read what you have typed here.
 
bobisaka said:
View attachment 28428
answer:
View attachment 28429


My solution:

Apply Quotient Property rule..

((25m^1/6 - ^4/6)/(n^-1/6 -11/6))^1/2

((25m^-3/6)/(n^12/6 ))^1/2

((25m^-1/2)/(n^2/1 ))^1/2

I believe I am stuck at this point, however I will do what i think...

((25)^1/2 (m^-1/2)^1/2)/(n^2/1)^1/2) SO FAR SO GOOD!

Simplify.. square integer & Apply product to power rule.. Power of 1/2 means square root not square

((sqr 25) (m^-1/2)^1/2)/((n^2/1)^1/2) <----- still unsure when to use square root vs 25*1/2, which equals 12.5 25^(1/2) is NOT the same as 25 * (1/2)
Click to expand...

bobisaka said:
((5) (m^-2/2)/(n^3/2)) You are not using the rule (a^m)^n = a^(m*n) correctly. You multiply the indices in this rule.
Click to expand...
-1/2 * 1/2 = -1/4 for the power of m
and
2/1 * 1/2 = 1 for the power of n

So you have 5 m^(-1/4) n^1 = (5 n) / (m^(1/4) where m^(-1/4) on the numerator becomes m^(1/4) on the denominator.


bobisaka said:
(5m)/(n^3/2)


And there you have, still lost as usual. I need to understand what I am doing wrong, before i can continue to next chapter..

How did 5 combine with the 'n' variable, when 25 was connected to 'm'.. so lost.
Click to expand...
I think your problem is multiplying fractions!

Have a read through all my comments in red next to your working.
 
Although people have been too kind to say so, part of your problem is your careless notation. It confuses us even though we understand the problem. I admit, however, that typing this kind of problem in line is a nightmare. Do it in paper and pencil, photograph it, and attach an image to your post

Let’s do this step by step. Work within grouping symbols first ALWAYS.

[math]\left ( \dfrac{25m^{1/6}n^{11/6}}{m^{2/3}n^{-1/6}} \right )^{1/2} = \left ( 25 * \dfrac{m^{1/6}}{m^{2/3}} * \dfrac{n^{11/6}}{n^{-1/6}} \right )^{1/2}[/math].

I isolated the numerals and each variable to let myself work on smaller, easier problems.

So here is a general rule that is relevant (I do not understand the current teaching obsession with teaching the names of properties). Otis already pointed this out.

[math]\dfrac{p^q}{p^r} = p^{(q - r)}.[/math]
We have two fractions like that, one in m and one in n.

[math]\dfrac{1}{6} - \dfrac{2}{3} = \dfrac{1}{6} - \dfrac{4}{6} = - \dfrac{3}{6} = -\dfrac{1}{2} \implies \dfrac{m^{1/6}}{m^{2/3}} = m^{-1/2}.[/math]
[math]\dfrac{11}{6} - \dfrac{-1}{6} = \dfrac{11}{6} + \dfrac{1}{6} = \dfrac{12}{6} = 2 \implies \dfrac{n^{11/6}}{n^{-1/6}} = n^2.[/math]
[math]\therefore \left ( 25 * \dfrac{n^{1/6}}{n^{2/3}} * \dfrac{m^{11/6}}{m^{-1/6}} \right )^{1/2} = (25 * m^{-1/2} * n^2)^{1/2} = (5^2 * n^{-1/2} * n^2)^{1/2}[/math].

Things look a lot simpler. Here is another general rule that applies

[math](a * b * c)^d = a^d * b^d * c^d.[/math]
[math]\therefore (5^2 * m^{-1/2} * n^2)^{1/2} = (5^2)^{1/2} * (m^{-1/2})^{1/2} * (n^2)^{1/2} = 5^1 * m^{-1/4} * n^1.[/math]
Can you finish now?
 
Hi all, thanks for your inputs. Yes, I suck pretty bad at explaining my thoughts. :D

JeffM, thanks for your post above, helped out alot.

Now I have worked it out on a piece of paper like you guys said.
I've noted down some points for the steps I was confused about in the past, but have now resolved (i think). There are also some questions in regards to the final answer which I am confused about, specifically how did step 8 turn into a fraction.

Please have a look at the attached image. Bit blurry on the left side.
 

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bobisaka said:
Hi all, thanks for your inputs. Yes, I suck pretty bad at explaining my thoughts. :D

JeffM, thanks for your post above, helped out alot.

Now I have worked it out on a piece of paper like you guys said.
I've noted down some points for the steps I was confused about in the past, but have now resolved (i think). There are also some questions in regards to the final answer which I am confused about, specifically how did step 8 turn into a fraction.

Please have a look at the attached image. Bit blurry on the left side.
Click to expand...
Step 8:
1. Identify the base and the exponent: the base is m, exponent is -1/4.
2. The negative sign of the exponent tells us that we are _dividing_ by the base raised to the _positive_ exponent:
[math]a^{-n}=\frac{1}{a^n}[/math]
So, [math]m^{-1/4}=\frac{1}{m^{1/4}}[/math]
As for 5n, we are multiplying by 5 and by n, so they end up in the numerator of the resulting fraction.
More: http://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html
 
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JeffM said:
… the current teaching obsession with teaching the names of properties …
Click to expand...
It has a name? (I must've forgotten.)

"A property of exponents for simplifying a ratio of powers having the same base"

Oh, wait. I think I understand the obsession for "nicknames". My moniker is way too long for mathematicians to stomach (being a naturally-lazy lot).

;)
 
Otis said:
It has a name? (I must've forgotten.)

"A property of exponents for simplifying a ratio of powers having the same base"

Oh, wait. I think I understand the obsession for "nicknames". My moniker is way too long for mathematicians to stomach (being a naturally-lazy lot).

;)
Click to expand...
I am not just naturally lazy. I have cultivated that innate disposition.
 
excellent.

Negative exponent m^-1/4 becomes a quotient where the negative sign becomes a numerator of positive 1 and the m^1/4 becomes the denominator.
The remaining 5n multiplies with the 1 numerator above the denominator.
The equation cannot be further simplified thus left as 5n/m^1/4.

My question now is, why does the negative sign becoming positive numerator? Is that just the rule? What is the explanation for doing so?
I guess i could ask the same thing as to why is a negative property become a quotient.. Anyways just trying to deepen my understanding of how it works.
 
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