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

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

Mathematics
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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.
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
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 \[3^5\cdot3^{4n}\]
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 \[3^{5n}=3^{5+4n}\]
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?
wait im confused like if its (3xy)^3 then the "3" would be 27 but there the 3 is still "3" whys that?
In the case of \((3xy)^3\), you can think of it as \[(3xy)(3xy)(3xy)=(3\cdot3\cdot3)\cdot(x\cdot x\cdot x)\cdot(y\cdot y\cdot y)=3^3\cdot x^3\cdot y^3=27x^3y^3\]
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 \[(3^{2n+1}) on the \right side im confused for the \left side\]
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.
*3^2n+2
Wait, on the right side, you have\[\Large (3^{n+1})^3\]What you gave, was a simplification for \[\Large (3^{n+1})^2\]
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
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 \[\large (3^2)^{2n}?\]
its 3^4n
Perfect. Now what is \[\large 3\cdot3^{4n}?\]
3^4n+1?
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.
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
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 ?
You may want to start a new question for each question to make it easier to read for others later on.
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
\[3^{4n}\] 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: \[81^{n}\]
3 has a degree of 0, huh?
in the degree of a monomial the constant has a degree of 0 right?
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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