anonymous
  • anonymous
use the exponent laws to simplify the following until you have only a single power. then evaluate.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
i dont even know what demoninator and numerator mean

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anonymous
  • anonymous
|dw:1327953048715:dw|
anonymous
  • anonymous
|dw:1327953037368:dw|
myininaya
  • myininaya
Here are some laws to know: \[\frac{x^a}{x^b}=x^{a-b} ; x^0=1 , x \neq 0\]
anonymous
  • anonymous
i know the laws i just dont know how to apply them.
myininaya
  • myininaya
\[\frac{2^4 \cdot (4^{5-2})}{1}=2^4 \cdot 4^3\]
myininaya
  • myininaya
So that should help
myininaya
  • myininaya
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! :)
anonymous
  • anonymous
apparently im im supost to make the 2 into another number
anonymous
  • anonymous
ugh!!! this is so confusing! !!!!!
myininaya
  • myininaya
Could you tell me where you are having the problem?
anonymous
  • anonymous
where? like the whole problem is a problem i dont know where to start or what to do
anonymous
  • anonymous
im homeschooled i dont have a teacher to teach me this stuff
myininaya
  • myininaya
well right a way i seen the bottom was one since something^0=1 (of course when that something isn't 0)
myininaya
  • myininaya
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
myininaya
  • myininaya
so that is how i got \[\frac{2^4 \cdot (4^{5-2})}{1}\]
myininaya
  • myininaya
\[2^4 \cdot (4^3)\]
myininaya
  • myininaya
since 5-2=3 and anything over 1 is that anything
anonymous
  • anonymous
but what happens to the stuff inside the brackets?
anonymous
  • anonymous
and apparently i have to change the 2 so its the same base or sumthin
myininaya
  • myininaya
\[2^4 \cdot 4^3\] you don't really need those parathesis anymore
anonymous
  • anonymous
ok i need to simplify.
myininaya
  • myininaya
oh you want to write it as 2^something?
anonymous
  • anonymous
i dont even know what you mean!
myininaya
  • myininaya
2 to some power?
myininaya
  • myininaya
\[2^{something}\] is that what you are saying what you want?
anonymous
  • anonymous
omg this is hopeless
myininaya
  • myininaya
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 \[2^4 \cdot 4^3\] \[16 \cdot 64\]
anonymous
  • anonymous
ok look at the original question
anonymous
  • anonymous
im supost to simplify that whole thing
myininaya
  • myininaya
so you wanted it as 2^something
anonymous
  • anonymous
with the same bases
myininaya
  • myininaya
\[2^4 \cdot (2^2)^3\] since 4 is 2(2)=2^2
myininaya
  • myininaya
\[2^4 \cdot 2^6\] since 2(3)=6
anonymous
  • anonymous
i dont even know what u mean..
anonymous
  • anonymous
omggggggggg!!!!!!
myininaya
  • myininaya
\[2^{4+6}\] using the law \[x^a \cdot x^b=x^{a+b}\]
myininaya
  • myininaya
i'm sorry can you tell me what is troubling you?
anonymous
  • anonymous
thanks for your help but i dont understand any math
anonymous
  • anonymous
i dont understand what to do or anything
anonymous
  • anonymous
i need like a step by step explination
myininaya
  • myininaya
All I'm doing is applying the law of exponents That is what I'm giving you
myininaya
  • myininaya
But if you don't understand the step let me know
anonymous
  • anonymous
i dont
myininaya
  • myininaya
ok which one?
anonymous
  • anonymous
all of it !
myininaya
  • myininaya
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
myininaya
  • myininaya
\[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0} \]
myininaya
  • myininaya
Ok on the bottom do you see how we can apply law 2?
anonymous
  • anonymous
your supost to do brackets first tho
myininaya
  • myininaya
It doesn't matter about all that we know the inside is not zero so we can apply law 2
anonymous
  • anonymous
so its 1? like for sure?
anonymous
  • anonymous
how do you know?
myininaya
  • myininaya
That is right ! Great Job! so we have \[2^{4} \times(4^{5}\div4^{2}) \over 1\] By law 2
myininaya
  • myininaya
\[2^{4} \times(4^{5}\div4^{2}) \]
myininaya
  • myininaya
So now this is what is left over
anonymous
  • anonymous
but u have to change the 2
myininaya
  • myininaya
By order of operations we are suppose to look inside the parathesis
myininaya
  • myininaya
What law can we apply inside the parathesis?
myininaya
  • myininaya
We have division and the same base so we can use law 1 is what i called it above
myininaya
  • myininaya
\[2^4 \times (4^{5-2})\] remember this by law 1
myininaya
  • myininaya
since 5-2=3 we can write \[2^4 \times 4^3\]
myininaya
  • myininaya
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 \[2^4 \times(2^2)^3\]
myininaya
  • myininaya
i replaced 4 with 2^2 since 4=2^2
anonymous
  • anonymous
i get all of it up to thr last bit
myininaya
  • myininaya
\[2^2=4?\]
anonymous
  • anonymous
im at 2^(4) x 4^(3)
myininaya
  • myininaya
ok what about 2^4 * 4^3?
anonymous
  • anonymous
im at that step
myininaya
  • myininaya
ok let me know if you have a question about that step
anonymous
  • anonymous
i have a question, like whats the next step after it
anonymous
  • anonymous
how do we change the 2 into a 4
myininaya
  • myininaya
this is what i wrote \[2^4 \times(2^2)^3 \]
myininaya
  • myininaya
\[\text{ remember } 2^2=4\]
anonymous
  • anonymous
whats the mini number
myininaya
  • myininaya
you mean the exponent?
anonymous
  • anonymous
sure i guess?
myininaya
  • myininaya
|dw:1327954725191:dw|
myininaya
  • myininaya
That is a two.
anonymous
  • anonymous
so after my last step what do i write lol
myininaya
  • myininaya
\[2^4 \times (2^2)^3\] 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 \[2^4 \cdot 2^6\]
anonymous
  • anonymous
what does the dot mean
myininaya
  • myininaya
multiply
myininaya
  • myininaya
Do you know what to do last?
anonymous
  • anonymous
k ima draw out what i have and u tell me if its correct ok?
anonymous
  • anonymous
|dw:1327955081274:dw|
anonymous
  • anonymous
|dw:1327955246139:dw|
anonymous
  • anonymous
|dw:1327955289855:dw|
anonymous
  • anonymous
|dw:1327955357557:dw|
anonymous
  • anonymous
|dw:1327955389263:dw|
anonymous
  • anonymous
|dw:1327955419197:dw|
anonymous
  • anonymous
|dw:1327955435698:dw|
anonymous
  • anonymous
is that it?
myininaya
  • myininaya
that is right
anonymous
  • anonymous
thank you so much! omg ! :)

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