Question

Pic Below Its An Essay Will Medal And Fan for Best Answer!

Answer 1

@Michele_Laino

Answer 2

@dan815 @Luigi0210

Answer 3

first step:
since \(9^0=1\) the I can write this:
\[\huge \begin{gathered}
{\left( {{3^8} \cdot {2^{ - 5}} \cdot {9^0}} \right)^{ - 2}} = {\left( {{3^8} \cdot {2^{ - 5}} \cdot 1} \right)^{ - 2}} = \hfill \\
\hfill \\
= {\left( {{3^8} \cdot {2^{ - 5}}} \right)^{ - 2}} = {3^{8 \cdot \left( { - 2} \right)}} \cdot {2^{ - 5 \cdot \left( { - 2} \right)}} = ...? \hfill \\
\end{gathered} \]

Answer 4

oops.. I rewrite such expression:
\[\Large \begin{gathered}
{\left( {{3^8} \cdot {2^{ - 5}} \cdot {9^0}} \right)^{ - 2}} = {\left( {{3^8} \cdot {2^{ - 5}} \cdot 1} \right)^{ - 2}} = \hfill \\
\hfill \\
= {\left( {{3^8} \cdot {2^{ - 5}}} \right)^{ - 2}} = {3^{8 \cdot \left( { - 2} \right)}} \cdot {2^{ - 5 \cdot \left( { - 2} \right)}} = ...? \hfill \\
\end{gathered} \]

Answer 5

please apply these identities:
\[\Large {\left( {{3^m}} \right)^n} = {3^{m \cdot n}},\quad {\left( {{2^k}} \right)^q} = {2^{k \cdot q}}\]
where \(m,\;n,\;k,\;q\) are integers

Answer 6

@Michele_Laino Sorry i had to go somewhere

Answer 7

I'm very sorry I have to go now, since I have to give a tutoring lesson here at my home. I will return between 2 hours

Answer 8

ok

Answer 9

thanks!! :)

Answer 10

@dan815 @nincompoop @Hero