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its the first one 5n is on three

\[3^{5n}=3^5\cdot(3^{2n})^2\]Correct?

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\[3^{5n}=3^{5}(3^{2n})^{2} solve for n\]

yes

oh hey can u help me on this prob? \[3*9^{2n}=(3^{n+1})^{3}\]

First off, write \[3\cdot9^{2n} =3\cdot(3^2)^{2n}\]Can you show the next step in simplifying this?

ok um its \[(3^{2n+1}) on the \right side im confused for the \left side\]

*3^2n+2

oh yeah my bad yeah its 3

please i get the right side is 3^3n+3 but what i do to the rigleft side?

left side

its 3^4n

Perfect. Now what is \[\large 3\cdot3^{4n}?\]

3^4n+1?

oh thank you very much hey one more

quick question

why is 3 in (3xy)^3 is 27... but 3in (3^2n)^2 is still 3
?

Is that \[(3^{2n})^{2}\] ?

yea

is it because we treat 3 as an x? so 3 doesnt change?

and also usually the 3 has a degree of 0 but here it is one

3 has a degree of 0, huh?

in the degree of a monomial the constant has a degree of 0 right?

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