3^5n=3^5(3^2n)^2 solve for n

- anonymous

3^5n=3^5(3^2n)^2 solve for n

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- anonymous

Please rewrite the question using the equation tool, as it's tough to tell if you mean:
\[3^{5n}\] or \[3^{5}n\] for example. Same thing goes for the other side of the equation.

- anonymous

its the first one 5n is on three

- KingGeorge

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

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## More answers

- anonymous

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

- anonymous

yes

- KingGeorge

First, you need to simplify the expression on the right side. Recall that \[(a^b)^c=a^{bc}\]This means that you can rewrite the right hand side as \[3^5\cdot3^{4n}\]

- KingGeorge

Now we also have the rule that \[a^b\cdot a^c=a^{b+c}\]That means we can rewrite the RHS again. This time as \[3^{5+4n}.\]That means we have the relation \[3^{5n}=3^{5+4n}\]

- KingGeorge

Since 3 is the base on both sides, we can just get rid of it (otherwise known as taking the log base 3 of both sides). Hence, we only have to solve the equation \[5n=5+4n\]Can you do this yourself?

- anonymous

wait im confused like if its (3xy)^3 then the "3" would be 27 but there the 3 is still "3" whys that?

- KingGeorge

In the case of \((3xy)^3\), you can think of it as \[(3xy)(3xy)(3xy)=(3\cdot3\cdot3)\cdot(x\cdot x\cdot x)\cdot(y\cdot y\cdot y)=3^3\cdot x^3\cdot y^3=27x^3y^3\]

- anonymous

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

- KingGeorge

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

- anonymous

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

- anonymous

It may help you if you temporarily replace the exponents with simpler variables. For example, you could say 2n = X, and n+1 = Y.
That way you can rewrite that equation as:
\[3*9^{x} = (3^{y})^{3}\]
From there you can use the rules George posted.

- anonymous

*3^2n+2

- KingGeorge

Wait, on the right side, you have\[\Large (3^{n+1})^3\]What you gave, was a simplification for \[\Large (3^{n+1})^2\]

- anonymous

oh yeah my bad yeah its 3

- anonymous

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

- anonymous

left side

- KingGeorge

For the left side, we have \[\large 3\cdot(3^2)^{2n}\]Using the same principles you just used for the right side, what is \[\large (3^2)^{2n}?\]

- anonymous

its 3^4n

- KingGeorge

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

- anonymous

3^4n+1?

- KingGeorge

Right again. That means we can take log base 3 of both sides, and get \[4n+1=3n+3\]From here, it's just a simple solving of a linear equation.

- anonymous

FYI if you do substitute 2n = X, and n+1 = Y:
\[3*9^{x} = 3^{3y}\]
Since
\[9 = 3^{2}\]
\[3*(3^{2})^{x} = 3^{3y}\]
\[3*3^{2x} = 3^{3y}\]
\[3^{1}*3^{2x} = 3^{3y}\]
\[3^{2x+1} = 3^{3y}\]
Take the log base 3 of both sides to get:
2x+1 = 3y
Replace x and y with their values to get:
2(2n)+1 = 3(n+1)
4n+1 = 3n+3

- anonymous

oh thank you very much hey one more

- anonymous

quick question

- anonymous

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

- anonymous

You may want to start a new question for each question to make it easier to read for others later on.

- anonymous

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

- anonymous

yea

- anonymous

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

- anonymous

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

- anonymous

\[3^{4n}\]
Because you don't know what the full exponent is, it's easier to keep it as it is.
You could rewrite it as
\[(3^{4})^{n}\]
and then solve for 3^4, so you'd get:
\[81^{n}\]

- anonymous

3 has a degree of 0, huh?

- anonymous

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

- KingGeorge

It's not the constant that has a degree, it's the variable that has degree 0. Because the degree is 0, the variable can be replaced with \(x^0=1\), so you see a constant term.

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