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

- anonymous

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

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- anonymous

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

- 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

i dont even know what demoninator and numerator mean

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## More answers

- anonymous

|dw:1327953048715:dw|

- anonymous

|dw:1327953037368:dw|

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

So that should help

- 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

apparently im im supost to make the 2 into another number

- anonymous

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

- myininaya

Could you tell me where you are having the problem?

- anonymous

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

- anonymous

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

- myininaya

well right a way i seen the bottom was one since something^0=1
(of course when that something isn't 0)

- 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

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

- myininaya

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

- myininaya

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

- anonymous

but what happens to the stuff inside the brackets?

- anonymous

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

- myininaya

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

- anonymous

ok i need to simplify.

- myininaya

oh you want to write it as 2^something?

- anonymous

i dont even know what you mean!

- myininaya

2 to some power?

- myininaya

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

- anonymous

omg this is hopeless

- 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

ok
look at the original question

- anonymous

im supost to simplify that whole thing

- myininaya

so you wanted it as 2^something

- anonymous

with the same bases

- myininaya

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

- myininaya

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

- anonymous

i dont even know what u mean..

- anonymous

omggggggggg!!!!!!

- myininaya

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

- myininaya

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

- anonymous

thanks for your help but i dont understand any math

- anonymous

i dont understand what to do or anything

- anonymous

i need like a step by step explination

- myininaya

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

- myininaya

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

- anonymous

i dont

- myininaya

ok which one?

- anonymous

all of it !

- 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

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

- myininaya

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

- anonymous

your supost to do brackets first tho

- myininaya

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

- anonymous

so its 1? like for sure?

- anonymous

how do you know?

- myininaya

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

- myininaya

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

- myininaya

So now this is what is left over

- anonymous

but u have to change the 2

- myininaya

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

- myininaya

What law can we apply inside the parathesis?

- myininaya

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

- myininaya

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

- myininaya

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

- 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

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

- anonymous

i get all of it up to thr last bit

- myininaya

\[2^2=4?\]

- anonymous

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

- myininaya

ok what about 2^4 * 4^3?

- anonymous

im at that step

- myininaya

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

- anonymous

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

- anonymous

how do we change the 2 into a 4

- myininaya

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

- myininaya

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

- anonymous

whats the mini number

- myininaya

you mean the exponent?

- anonymous

sure i guess?

- myininaya

|dw:1327954725191:dw|

- myininaya

That is a two.

- anonymous

so after my last step what do i write lol

- 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

what does the dot mean

- myininaya

multiply

- myininaya

Do you know what to do last?

- anonymous

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

- anonymous

|dw:1327955081274:dw|

- anonymous

|dw:1327955246139:dw|

- anonymous

|dw:1327955289855:dw|

- anonymous

|dw:1327955357557:dw|

- anonymous

|dw:1327955389263:dw|

- anonymous

|dw:1327955419197:dw|

- anonymous

|dw:1327955435698:dw|

- anonymous

is that it?

- myininaya

that is right

- anonymous

thank you so much! omg ! :)

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