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Mathematics 16 Online
OpenStudy (volleyballlover55):

help!

OpenStudy (volleyballlover55):

OpenStudy (volleyballlover55):

@TheSmartOne @rebeccaxhawaii

OpenStudy (volleyballlover55):

@Jakesaurus1127

OpenStudy (volleyballlover55):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

What do you have so far?

OpenStudy (volleyballlover55):

nothing..

OpenStudy (volleyballlover55):

@jim_thompson5910 ???

jimthompson5910 (jim_thompson5910):

Have you heard of the rule that \[\LARGE \left(x^y\right)^z = x^{y*z}\] ?

OpenStudy (volleyballlover55):

yes

jimthompson5910 (jim_thompson5910):

So we'll use this rule to break down the expression

OpenStudy (volleyballlover55):

ok!

jimthompson5910 (jim_thompson5910):

For instance \[\LARGE \left(3^{8}*2^{-5}*9^{0}\right)^{-2} = 3^{8*(-2)}*2^{-5*(-2)}*9^{0*(-2)}\] Notice how the outer exponent (-2) is multiplied by EVERY exponent inside. It's kinda like distributing.

jimthompson5910 (jim_thompson5910):

make sense?

OpenStudy (volleyballlover55):

yea

jimthompson5910 (jim_thompson5910):

what is 8*(-2) equal to?

OpenStudy (volleyballlover55):

-16

jimthompson5910 (jim_thompson5910):

how about -2*(-5) ?

OpenStudy (volleyballlover55):

10

jimthompson5910 (jim_thompson5910):

So this means \[\Large \left(3^{8}*2^{-5}*9^{0}\right)^{-2} = 3^{8*(-2)}*2^{-5*(-2)}*9^{0*(-2)}\] \[\Large \left(3^{8}*2^{-5}*9^{0}\right)^{-2} = 3^{-16}*2^{10}*9^{0}\] \[\Large \left(3^{8}*2^{-5}*9^{0}\right)^{-2} = 3^{-16}*2^{10}*1\] \[\Large \left(3^{8}*2^{-5}*9^{0}\right)^{-2} = 3^{-16}*2^{10}\]

jimthompson5910 (jim_thompson5910):

Do you agree?

OpenStudy (volleyballlover55):

yes!

jimthompson5910 (jim_thompson5910):

ok now onto the piece \[\Large \left(\frac{2^{-2}}{3^{3}}\right)^{4}\]

jimthompson5910 (jim_thompson5910):

The \(\Large 3^3\) in the denominator can be written as \(\Large 3^{-3}\) In other words, \[\Large \frac{1}{3^3} = 3^{-3}\] do you agree?

OpenStudy (jakesaurus1127):

@volleyballlover55 the answer to your other problem was \[(3^{8}*2^{-5}*9^{0})^{-2}*(\frac{2^{-2}}{3^{3}})^{4}*3^{28}\]\[(6^{3})^{-2}*(\frac{.25}{27})^{4}*22,876,792,454,961\]\[0.0000214335*0.0092592592*22,876,792,454,961\]\[4540090.07357\] sorry it took so long.

OpenStudy (volleyballlover55):

isnt the answer 4?

OpenStudy (volleyballlover55):

@jim_thompson5910 is he correct?

jimthompson5910 (jim_thompson5910):

do you agree with what I posted last @volleyballlover55 ?

OpenStudy (volleyballlover55):

yes..

jimthompson5910 (jim_thompson5910):

So, \[\Large \left(\frac{2^{-2}}{3^{3}}\right)^{4}=\left(2^{-2}*3^{-3}\right)^{4}\]

jimthompson5910 (jim_thompson5910):

What do you get when you multiply the outer exponent (4) by the inner exponents?

OpenStudy (volleyballlover55):

um, idk..

jimthompson5910 (jim_thompson5910):

first tell me what the inner exponents are

jimthompson5910 (jim_thompson5910):

any ideas @volleyballlover55 ?

OpenStudy (volleyballlover55):

um.. 531441/256

jimthompson5910 (jim_thompson5910):

I'd leave it in exponent form

jimthompson5910 (jim_thompson5910):

example: instead of saying 1024, write 2^10

jimthompson5910 (jim_thompson5910):

do you know what I mean by "exponent form" ?

OpenStudy (volleyballlover55):

yes..

jimthompson5910 (jim_thompson5910):

Use the rule \[\Large \left(x^y\right)^z = x^{y*z}\] and tell me what \[\Large \left(2^{-2}\right)^{4}\] is equal to in exponent form

OpenStudy (volleyballlover55):

1/256

jimthompson5910 (jim_thompson5910):

the exponents are -2 and 4. Multiply them to get -8 so \[\Large \left(2^{-2}\right)^{4}=2^{-2*4}=2^{-8}\] That's what I meant by "exponent form"

jimthompson5910 (jim_thompson5910):

Let's try another for practice. What is (3^5)^7 in exponent form?

OpenStudy (volleyballlover55):

can we please just get onto the problem.. i need to finish it..

jimthompson5910 (jim_thompson5910):

If you aren't going to work with me, then I can't help you. Sorry

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