"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

 

[PeroK]

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

One of the reasons that this notation is useful is that many of the natural constants and quantities are either very large or very small in SI units. In physics, we have things like:

The mass of an electron is $9.1 \times 10^{-31} kg$; and the mass of the Sun is $2 \times 10^{30}kg$.

It would be very confusing to write these out with a long line of thirty zeroes.

 

[chriscarson]

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

you have to know these things well before try to solve this sums. i ll see how it goes with my next work . thanks

 

[chriscarson]

One of the reasons that this notation is useful is that many of the natural constants and quantities are either very large or very small in SI units. In physics, we have things like:

The mass of an electron is $9.1 \times 10^{-31} kg$; and the mass of the Sun is $2 \times 10^{30}kg$.

It would be very confusing to write these out with a long line of thirty zeroes. and you can see the digits how

that what standard form are used for from what I see. from an electron to the sun there is an outstanding gap of size but still, the can be written in a little amount of digits.

 

[Delta2]

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

It is important to notice that the equation $$10^n\cdot 10^m=10^{n+m}$$ holds even when m and n are negative numbers (or when one of n, m is negative and the other positive). For example $$\frac{10^3}{10^5}=10^3\cdot 10^{-5}=10^{3+(-5)}=10^{-2}=\frac{1}{10^2}$$

 

[etotheipi]

Plus there is a natural correspondence between the naming system and the standard form, e.g. two nonillion, nine hundred and eleven decillionths?

Also