(2^8 ⋅ 3^−5 ⋅ 6^0)^−2 ⋅ 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 2^28 (5 points)

Question

anonymous · Answer

(2^8 x 3^5 x 6^0)^-2 x 3^-2/2^3 x 2^28

igreen · Answer

Is this it? @cami1231

$\dfrac{(2^8 ⋅ 3^5 ⋅ 6^0)^{-2} ⋅ 3^{-2}}{2^3 ⋅ 2^{28}}$

anonymous · Answer

$$(2^{8}*3^{5}*6^{0})^{-2} *3^{-2}/2^{3}*2^{28}$$

anonymous · Answer

I think so! Thanks so much! @igreen @juniro

anonymous · Answer

i might have other questions tho !

anonymous · Answer

(256*243*1)^-2*(1/72)*2^28

=512/531441=.000963

anonymous · Answer

actually i am wrong i went of your written equation and you missed a negative on 3^5 so your final answer should be 3359232

anonymous · Answer

Thanks @juniro

anonymous · Answer

oy yeah its 3^-5

igreen · Answer

$\dfrac{(2^8 ⋅ 3^5 ⋅ 6^0)^{-2} ⋅ 3^{-2}}{2^3 ⋅ 2^{28}}$

Simplify the denominator by adding the exponents and keeping the base:

$\dfrac{(2^8 ⋅ 3^5 ⋅ 6^0)^{-2} ⋅ 3^{-2}}{2^{31}}$

Multiply -2 to all the exponents that are in the parenthesis:

$\dfrac{2^{-16} ⋅ 3^{-10} ⋅ 6^0 ⋅ 3^{-2}}{2^{31}}$

Simlify 3^-10 and 3^-2 by adding the exponents and keeping the base:

$\dfrac{2^{-16} ⋅ 3^{-12} ⋅ 6^0 }{2^{31}}$

Simplify 6^0:

$\dfrac{2^{-16} ⋅ 3^{-12}}{2^{31}}$

That's all you can simplify it without simplifying the exponents.

anonymous · Answer

Thank you ! @igreen but i made an error when typing.
its supposed to be 3^-5

igreen · Answer

That's okay:

$\dfrac{2^{-16} ⋅ 3^{10} ⋅ 6^0 ⋅ 3^{-2}}{2^{31}}$

Simplify 6^0:

$\dfrac{2^{-16} ⋅ 3^{10} ⋅ 3^{-2}}{2^{31}}$

Simlify 3^10 and 3^-2 by adding the exponents and keeping the base:

$\dfrac{2^{-16} ⋅ 3^8}{2^{31}}$

anonymous · Answer

thank you ! @igreen ♥

igreen · Answer

No problem!