use the exponent laws to simplify the following until you have only a single power. then evaluate.

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use the exponent laws to simplify the following until you have only a single power. then evaluate.

Mathematics
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\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}\]
ok sarah, first, (4^5)/(4^2)=4^3 4^3=2^4, so we have (2^4)*(2^4) in the denominator, but i can't see the numenator clearly, can u type it again
i dont even know what demoninator and numerator mean

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Other answers:

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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
well right a way i seen the bottom was one since something^0=1 (of course when that something isn't 0)
then the two guys inside that parenthesis had the same base and i was like we should apply this law \[\frac{x^a}{x^b}=x^{a-b}\] since we had the same base and we are dividing
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
i was just going to write 2^4 as 2(2)(2)(2)=16 and 4^3=4(4)(4)=64 and find the product of 16 and 64 since we have \[2^4 \cdot 4^3\] \[16 \cdot 64\]
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 !
ok so I'm going to write down all the law of exponents we will need: Law 1:\[\frac{x^a}{x^b}=x^{a-b}\] Law 2:\[x \neq 0 , x^0=1\] Law 3:\[x^a \cdot x^b=x^{a+b}\] Law 4:\[(x^a)^b=x^{ a \cdot b}\] These are the laws I'm going to use
\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0} \]
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 \[2^4 \times 4^3\]
now since we want the same base we need find out how to write 4 such that it equals 2 to some power \[4=2^2 \] so we can write \[2^4 \times(2^2)^3\]
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
\[2^4 \times (2^2)^3\] We can rewrite that second part as \[2^{2 \cdot 3}=2^6\] by use of one of the other laws so now we are left with \[2^4 \cdot 2^6\]
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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is that it?
that is right
thank you so much! omg ! :)

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