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

[chriscarson]

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 , I am not getting the right answer.

 

[Delta2]

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 $10^m\times10^n=10^{m+n}$

 

[PeroK]

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]

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

 

[chriscarson]

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 .

 

[chriscarson]

now I remember something like this

 

[Delta2]

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 ?

 

[Delta2]

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 ,

 

[chriscarson]

THANKS ALL OF YOU AGAIN

 

[FactChecker]

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]

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]

These might be useful

https://mathinsight.org/exponentiation_basic_rules
https://www.uea.ac.uk/documents/6207125/8183824/steps+into+numeracy+powers+of+10+and+standard+form.pdf

 

[PeroK]

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]

These might be useful

https://mathinsight.org/exponentiation_basic_rules
https://www.uea.ac.uk/documents/6207125/8183824/steps+into+numeracy+powers+of+10+and+standard+form.pdf

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

 

[chriscarson]

$10^2 = 10 \times 10 = 100$. 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

 

[chriscarson]

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

 

[chriscarson]

$10^2 = 10 \times 10 = 100$. 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]

View attachment 265329

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

What do you mean it doesn't work for you, you haven't been introduced to negative numbers? or you don't know that $10^{-n}=\frac{1}{10^n}$

 

[PeroK]

so I had a bit of a good reason

Yes, but just be careful not to write $9^2 = 99$. That's the thing to avoid.

 

[PeroK]

View attachment 265329

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

This is the same, except the number on the denominator is bigger. So, you get:
$$\frac{10^7}{10^9} = 10^{-2} = \frac{1}{10^2} = \frac 1 {100}$$
Note that the last three expressions all mean exactly the same thing.

 

[chriscarson]

This is the same, except the number on the denominator is bigger. So, you get:
$$\frac{10^7}{10^9} = 10^{-2} = \frac{1}{10^2} = \frac 1 {100}$$
Note that the last three expressions all mean exactly the same thing.

that s a good explanation , I have to remember it now. Thanks

 

[chriscarson]

What do you mean it doesn't work for you, you haven't been introduced to negative numbers? or you don't know that $10^{-n}=\frac{1}{10^n}$

I was introduced but was nt sure were they can be used or sometimes forget the mechanisim.

 

[chriscarson]

Yes, but just be careful not to write $9^2 = 99$. That's the thing to avoid.

ok, at least I am sure that 9 to the power of 2 is 9x9.

 

[etotheipi]

that s a good explanation , I have to remember it now. Thanks

You could also note that it's consistent with the properties you already know. $x^n x^{-n} = x^{n-n} = x^0 = 1$, so divide through by $x^n$ and you see $x^{-n} = \frac{1}{x^n}$.

But N.B. AFAIK $x^0 := 1$ and $x^{-n} := \frac{1}{x^n}$ are definitions, and the exponent laws can be proven by induction. So the first sentence is deriving it backwards