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\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}\]

i dont even know what demoninator and numerator mean

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Here are some laws to know:
\[\frac{x^a}{x^b}=x^{a-b} ; x^0=1 , x \neq 0\]

i know the laws i just dont know how to apply them.

\[\frac{2^4 \cdot (4^{5-2})}{1}=2^4 \cdot 4^3\]

So that should help

and 2^4=2(2)(2)(2)=16
and 4^3=4(4)(4)=64
find the product of 64 and 16 and you will win! :)

apparently im im supost to make the 2 into another number

ugh!!! this is so confusing! !!!!!

Could you tell me where you are having the problem?

where? like the whole problem is a problem i dont know where to start or what to do

im homeschooled i dont have a teacher to teach me this stuff

so that is how i got
\[\frac{2^4 \cdot (4^{5-2})}{1}\]

\[2^4 \cdot (4^3)\]

since 5-2=3
and anything over 1 is that anything

but what happens to the stuff inside the brackets?

and apparently i have to change the 2 so its the same base or sumthin

\[2^4 \cdot 4^3\]
you don't really need those parathesis anymore

ok i need to simplify.

oh you want to write it as 2^something?

i dont even know what you mean!

2 to some power?

\[2^{something}\] is that what you are saying what you want?

omg this is hopeless

ok
look at the original question

im supost to simplify that whole thing

so you wanted it as 2^something

with the same bases

\[2^4 \cdot (2^2)^3\]
since 4 is 2(2)=2^2

\[2^4 \cdot 2^6\]
since 2(3)=6

i dont even know what u mean..

omggggggggg!!!!!!

\[2^{4+6}\]
using the law
\[x^a \cdot x^b=x^{a+b}\]

i'm sorry can you tell me what is troubling you?

thanks for your help but i dont understand any math

i dont understand what to do or anything

i need like a step by step explination

All I'm doing is applying the law of exponents
That is what I'm giving you

But if you don't understand the step let me know

i dont

ok which one?

all of it !

\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0} \]

Ok on the bottom
do you see how we can apply law 2?

your supost to do brackets first tho

It doesn't matter about all that
we know the inside is not zero
so we can apply law 2

so its 1? like for sure?

how do you know?

That is right ! Great Job! so we have
\[2^{4} \times(4^{5}\div4^{2}) \over 1\]
By law 2

\[2^{4} \times(4^{5}\div4^{2}) \]

So now this is what is left over

but u have to change the 2

By order of operations we are suppose to look inside the parathesis

What law can we apply inside the parathesis?

We have division and the same base so we can use law 1 is what i called it above

\[2^4 \times (4^{5-2})\]
remember this by law 1

since 5-2=3
we can write
\[2^4 \times 4^3\]

i replaced 4 with 2^2 since
4=2^2

i get all of it up to thr last bit

\[2^2=4?\]

im at 2^(4) x 4^(3)

ok what about 2^4 * 4^3?

im at that step

ok let me know if you have a question about that step

i have a question, like whats the next step after it

how do we change the 2 into a 4

this is what i wrote
\[2^4 \times(2^2)^3 \]

\[\text{ remember } 2^2=4\]

whats the mini number

you mean the exponent?

sure i guess?

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That is a two.

so after my last step what do i write lol

what does the dot mean

multiply

Do you know what to do last?

k ima draw out what i have and u tell me if its correct ok?

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|dw:1327955357557:dw|

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is that it?

that is right

thank you so much! omg ! :)