Question

what is the simplified form of the expression [m^-1 m^5/m^-2]^3

Answer 1

Do you know:

\[\large a^x \cdot a^y = a^{x + y}\]

Answer 2

In numerator, you have -1 and 5 in the powers of m..

When base is same(here base is m), then exponents add when you multiply...

Answer 3

Just add the powers in the numerator part...

Answer 4

These types of problems lose me... I dont understand why they put letters with numbers. im dyslexic so it makes it 100xs harder.. so bare with me.

Answer 5

I am here with you, don't worry..

Clearly tell me, where you are not getting, I will start from that point?

Answer 6

the negative numbers do i add them together or do i subtract the pos. from neg. or am i over thinking completely?

Answer 7

See, if you are adding or subtracting, there is no difference..

But don't interchange the sign anywhere..

You have -1 and 5 as powers..

Now you want to add them, so you will do like this : (-1) + (5)

Just see the addition sign in between them...

Answer 8

okay

Answer 9

so 6?

Answer 10

Now, just see that you have + sign with 5 and -ve sign with 1..

So, sign you will not interchange, yes you can interchange the terms but not the sign..

See here:

Can I write this (-1) + (5) as:

(5) + (-1)

Answer 11

okay so i can change the format it is, but dont change the sign at all.

Answer 12

See there, 5 is positive and - sign I have attached to 1..

And please note that if there is no sign before any number, that number is by default positive
I mean (5) = +(5) these both are one and the same..

Answer 13

ok i see that =]

Answer 14

Yes you can change the format, but not the sign..

Now let us see how to deal with situation where you have two signs:

See, you have + (-1) there, right??? (Forget about 5 for sometime)

Answer 15

Here you have two signs, first + and then - with 1.. Right??

Answer 16

right

Answer 17

So, just remember these rules:

+(-a) = -a (When + is outside and - is inside, then -will win)
-(+a) = -a (here also out of - outside and + inside, - will win)

-(-a) = +a (If you have two - signs, then it becomes +)
+(+a) = +a (+ and + remain +)

Answer 18

So, you have +(-1)

So now seeing above rules, try to tell me:

\(\large + (-1) = ??\)

Answer 19

-1 right?

Answer 20

Good...

Answer 21

And suppose you have : -(-1)

In this case what you will get?? (Just look above rules)

Answer 22

+1

Answer 23

So, I assume you are getting it correctly..

Now tell me can I write (5) + (-1) as (5) - 1 ??

Answer 24

no because the one is neg

Answer 25

Or Simply as 5 is positive so, there is no need to keep brackets with 5 :

\(\large 5 - 1 = ??\)

Answer 26

See, you have solved the last part right now..

you yourself told me that +(-1) = -1, Right??

Answer 27

yes

Answer 28

Then you can write -1 there now..

See I show you clearly, just look for red colored part:

\[ \large 5 \color{red}{+ (-1)} \implies 5 \color{red}{-1} = How \; \; much \; \; ??\]

Answer 29

4

Answer 30

-4

Answer 31

How minus??

Answer 32

5 is greater than 1..

So, sign will be of greater number... Since 5 has positive sign so, sign will also be positive...

Answer 33

ph ok yes i get it

Answer 34

oh*

Answer 35

See, you have 5 chocolates, you ate 1 chocolate, now how many chocolates you have??

Answer 36

4

Answer 37

\(5 -1 = 4\)
\(1 - 5 = -4\)

Getting why?

Answer 38

sure do =] finally

Answer 39

Here in 1 - 5, it means you have 1 chocolate, and you ate 5 chocolates, But HOW IT IS POSSIBLE..

So, here also, you will do 5 - 1, but as 5 has -ve sign, so, -ve sign you will give to final answer too... Hope you are getting it..

Answer 40

Coming back to our question:

In numerator:

You got :

\[\large m^{-1} \cdot m^5 = (m)^{\color{red}{-1 + 5}} = (m)^\color{red}{4}\]

Answer 41

Getting??

Answer 42

so in the red would it be me adding the -1+5 and the 4?

Answer 43

So, now we have:

\[\large (\frac{m^4}{m^{-2}})^3\]

Answer 44

I have added in the very next term, see there I have written 4..

Answer 45

See, we have solved somewhat, In numerator you have two m terms, now in numerator you have one m only..

Answer 46

right

Answer 47

Having difficulty anywhere till now, you must ask me right now...

Answer 48

no

Answer 49

So. let us proceed further:

Our second formula that we have to apply here is:

\[\huge \frac{a^x}{a^y} = a^{x - y}\]

Answer 50

See, you have one m in the numerator and now one m in the denominator, Right??

Answer 51

Please note: Earlier, you have two m in multiplication like (m.m) in this case, you have to add powers..

But when the two m will be in division like (m/m) then you will subtract the powers in this case..

Are you getting the difference between these two statements?

Answer 52

right

Answer 53

yes

Answer 54

So now you have powers as 4 in numerator and (-2) in denominator..

So, you will subtract them like this : Please note the red sign:

\[\large 4 \color{red}{-} (-2)\]\

Getting??

Answer 55

yes

Answer 56

PLEASE NOTE: This you must remember that numerator power will come first, then - sign in between and then denominator power..

Don't ever do this : \(\large (-2) \color{red}{-} (4)\) THIS IS WRONG..

Answer 57

So please be careful here...

Answer 58

Suppose you have : \(\huge \frac{m^{-2}}{m^4}\)

YES, IN THIS CASE YOU CAN SURELY DO : (-2) - (4) as numerator power is now (-2) and denominator power is 4..

Answer 59

okay

Answer 60

Now tell me in 4 - (-2)

\(- (-2) = ??\)

Answer 61

2

Answer 62

Or +2 right??

Answer 63

right

Answer 64

So, you can do like this now:

Look for the red part here also:

\[\large 4 \color{red}{- (-2)} \implies 4 \color{red}{+2} = ??\]

Answer 65

6

Answer 66

Yep..

Getting how we did it??

Answer 67

yeppers =]

Answer 68

So, are we left with now:

\[\large \color{blue}{(m^6)^3}\]

Answer 69

We have solved somewhat further, we had one m upside and one m downside, and now by solving that also, we have now one m...

Answer 70

right

Answer 71

Now third formula which you have to remember here which is simple also is:

\[\large \color{green}{(a^x)^y = a^{xy}}\]

In this case powers will get simply multiplied..

Answer 72

So, just try and tell me, can you reach your answer from here??

Answer 73

You are just one step away to reach there, just try otherwise I am here only..

Answer 74

Hey, just try and tell the answer, at the most you can be right only, nothing greater than that... :)

Answer 75

so the last part is to mul

Answer 76

Yep, go ahead..

Answer 77

6? i feel that is wrong though....

Answer 78

x = 6 and y = 3 that is outside, multiply them just..

Answer 79

18

Answer 80

Yes you are there..

but 18 is in the power of m...

Answer 81

so would it be 1/m18?

Answer 82

List of Fomulae that we have used here:


\[\large \color{green}{1. \; \; a^x \cdot a^y = a^{x+y}}\]
\[\large \color{red}{2. \; \; \frac{a^x}{a^y} = a^{x-y}}\]
\[\large \color{blue}{(a^x)^y = a^{x \cdot y}}\]

Answer 83

See, it will not go in denominator, as power is 18 and that is positive..

Answer 84

\(m^{18}\) will remain in numerator..

Yes if you would have -18 in the power, then it would be what you are thinking..

Answer 85

just m^18?

Answer 86

Yep..

Answer 87

\[\huge \color{green}{(m)^{-a} = \frac{1}{(m)^a}}\]

Answer 88

okay I get it that is what I was thinking it was at first but i wasnt sure because the problem had the (/) sign and i wasnt sure if i needed to keep it. thats is why i asked if it was 1/m18

Answer 89

but i think that you have helped me more than any math teacher thank you so much!

Answer 90

No \ sign we had removed when we did:

\[\large \color{green}{ \frac{m^4}{m^{-2}} = m^{4 + 2} = m^6}\]

Answer 91

See, since you are somewhat new to this topic, I am providing you whole solution step by step, Just go through it and learn how we did it...

Answer 92

We have our problem:

\(\huge (\frac{m^{-1} \cdot m^5}{m^{-2}})^3\)

\(\huge \implies (\frac{m^{-1 + 5}}{m^{-2}})^3\)
\(\huge \implies (\frac{m^4}{m^{-2}})^3\)
\(\huge \implies (m^{4 - (-2)})^3\)
\(\huge \implies (m^{4+2})^3\)
\(\huge \implies (m^6)^3\)
\(\huge \implies (m)^{6 \cdot 3}\)

\(\huge \color{blue}{\implies (m)^{18}}\)

Answer 93

If you have any doubt in any step, just let me know, but the main thing is, you should get each and every step, because these formulae and rules, which I have told you will be used frequently in next problems you will do...

So, try more and more problems, I am sure you will get familiar as well as comfortable with these...

Don't afraid of these, just make them your friends, they are real easy...

Answer 94

*be

Answer 95

can you check to see if i have the right answer on another one i just did on my own?

Answer 96

I can check only, hurry up because I have to have my dinner now..

Be quick..