Question

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

Answer 1

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.

Answer 2

its the first one 5n is on three

Answer 3

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

Answer 4

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

Answer 5

yes

Answer 6

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 \[3^5\cdot3^{4n}\]

Answer 7

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 \[3^{5n}=3^{5+4n}\]

Answer 8

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?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

*3^2n+2

Answer 16

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

Answer 17

oh yeah my bad yeah its 3

Answer 18

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

Answer 19

left side

Answer 20

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

Answer 21

its 3^4n

Answer 22

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

Answer 23

3^4n+1?

Answer 24

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.

Answer 25

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

Answer 26

oh thank you very much hey one more

Answer 27

quick question

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

yea

Answer 32

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

Answer 33

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

Answer 34

\[3^{4n}\]
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:
\[81^{n}\]

Answer 35

3 has a degree of 0, huh?

Answer 36

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

Answer 37

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.