# "To the power of" (powers in division have to be subtracted)

## Homework Statement:

can t figure out how to do this with out a calculator

## Relevant Equations:

powers in division have to be subtracted
Our teacher said that powers in division have to be subtracted when the same base , but still , im not getting the right answer.

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## Answers and Replies

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Delta2
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well you have a big mistake here it is not ##10\times 10^6=100^6## it is rather ##10\times 10^6=10^7##. To see this write ##10^6## as ##10\times10\times10\times10\times10\times10##. Generally it is ##10^m\times10^n=10^{m+n}##

chriscarson
PeroK
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One problem is that:

##10 \times 10^6 \ne 100^6##

Why not try doing the problem in longhand and check each step?

You start with ##\frac{10 \times 10^6}{10^5}## on one side of the page and ##\frac{10 \times 1,000,000}{100,000}## on the other side of the page.

Work through it both ways to see what is going on at each step.

chriscarson
well you have a big mistake here it is not ##10\times 10^6=100^6## it is rather ##10\times 10^6=10^7##. To see this write ##10^6## as ##10\times10\times10\times10\times10\times10##. Generally it is ##10^m\times10^n=10^{m+n}##
yeah that s true , I knew how 10 s you have to make but I asked if there is a short way to do by memory

jbriggs444
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2019 Award
The rules of precedence (PEMDAS or BEDMAS) call for exponentiation to be performed before multiplication or division. So it is interpreted as$$10 \times (10^6)$$ rather than as $$(10 \times 10)^6$$.

PEMDAS = Parentheses, Exponentiation, Multiplication/Division, Addition/Subtraction
BEDMAS = Brackets, Exponentiation, Division/Multiplication, Addition/Subtraction.

chriscarson
One problem is that:

##10 \times 10^6 \ne 100^6##

Why not try doing the problem in longhand and check each step?

You start with ##\frac{10 \times 10^6}{10^5}## on one side of the page and ##\frac{10 \times 1,000,000}{100,000}## on the other side of the page.

Work through it both ways to see what is going on at each step.
this is shorter though , but still you need to make a big sum , but it s better for sure

The rules of precedence (PEMDAS or BEDMAS) call for exponentiation to be performed before multiplication or division. So it is interpreted as$$10 \times (10^6)$$ rather than as $$(10 \times 10)^6$$.

PEMDAS = Parentheses, Exponentiation, Multiplication/Division, Addition/Subtraction
BEDMAS = Brackets, Exponentiation, Division/Multiplication, Addition/Subtraction.
Yes not because I remembered that but the multiplication I was doing first .

now I remember something like this

PeroK
Delta2
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yeah that s true , I knew how 10 s you have to make but I asked if there is a short way to do by memory
the short way is that ##10^m\times 10^n=10^{m+n}## Just apply this for ##m=1,n=6##.

chriscarson
the short way is that ##10^m\times 10^n=10^{m+n}## Just apply this for ##m=1,n=6##.
so when there is no power the power is 1 ?

jbriggs444
Delta2
Homework Helper
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so when there is no power the power is 1 ?
yes. It is ##10^1=10##

chriscarson
oh that s what I didn t know ,

Delta2
THANKS ALL OF YOU AGAIN

FactChecker
Gold Member
this is shorter though , but still you need to make a big sum , but it s better for sure
"make a big sum"? No. I think you may be doing something wrong.
6+1=7 gives ##10 * 10^6 = 10^7##. Then 7-5=2 gives ##\frac {10^7} {10^5} = 10^2 = 100##. So you should never have to deal with any "big sum".

chriscarson
"make a big sum"? No. I think you may be doing something wrong.
6+1=7 gives ##10 * 10^6 = 10^7##. Then 7-5=2 gives ##\frac {10^7} {10^5} = 10^2 = 100##. So you should never have to deal with any "big sum".
oh that s what exactly I was looking for , thanks the others helped me too in a way to understand what this means.

so easy for you with a high IQ

PeroK
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Gold Member
oh that s what I didn t know ,
##10^0 = 1## is the tricky one. That's the one you have to remember!

The other way to look at this is:
$$10 \times 10^6 = 10 \times (10 \times 10 \times 10 \times 10 \times 10 \times 10) = 10^7$$
And:
$$\frac{10^7}{10^5} = \frac{10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10}{ 10 \times 10 \times 10 \times 10 \times 10} = 10 \times 10 = 10^2$$
In the main step, we are cancelling five 10's on the bottom with five 10's on the top. That's where the "subtraction" rule comes from: ##7 - 5 = 2##.

chriscarson
##10^0 = 1## is the tricky one. That's the one you have to remember!

The other way to look at this is:
$$10 \times 10^6 = 10 \times (10 \times 10 \times 10 \times 10 \times 10 \times 10) = 10^7$$
And:
$$\frac{10^7}{10^5} = \frac{10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10}{ 10 \times 10 \times 10 \times 10 \times 10} = 10 \times 10 = 10^2$$
In the main step, we are cancelling five 10's on the bottom with five 10's on the top. That's where the "subtraction" rule comes from: ##7 - 5 = 2##.
and also it confused me that 10 when is in the power of 2 for example , the 0 with the 1 in the value of 10 is one of the two zeros you need to have , I thought you had to add another two zeros .

etotheipi
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2019 Award
chriscarson
PeroK
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and also it confused me that 10 when is in the power of 2 for example , the 0 with the 1 in the value of 10 is one of the two zeros you need to have , I thought you had to add another two zeros .
##10^2 = 10 \times 10 = 100##. In the same way that ##9^2 = 9 \times 9 = 81##.

Saying ##10^n## is a "one followed by ##n## zeroes" is true, but that's not the underlying rule.

chriscarson
oh thanks , I notice that many of them they have same bases that s what was like to search in youtube was. and then I have all sort of them with -, decimals etc like this

##10^2 = 10 \times 10 = 100##. In the same way that ##9^2 = 9 \times 9 = 81##.

Saying ##10^n## is a "one followed by ##n## zeroes" is true, but that's not the underlying rule.
I followed the rule of the decimal where is it and I thought in that case of 10 is behind the zero than add 2 spaces

so when is 10 ,20, 30 . etc is different

for example now it doesn t work for me 7 - 9

##10^2 = 10 \times 10 = 100##. In the same way that ##9^2 = 9 \times 9 = 81##.

Saying ##10^n## is a "one followed by ##n## zeroes" is true, but that's not the underlying rule.
so I had a bit of a good reason

Delta2
Homework Helper
Gold Member
View attachment 265329

for example now it doesn t work for me 7 - 9
What do you mean it doesnt work for you, you havent been introduced to negative numbers? or you dont know that ##10^{-n}=\frac{1}{10^n}##

chriscarson
PeroK