How do I get y on its own?

Formula is in an answer.

Question

How do I get y on its own?

Formula is in an answer.

anonymous · Answer

$$2(3^{3y})-5(3^{2y})-9(3^{y})$$

bobo-i-bo · Answer

$$2(3^{3y})-5(3^{2y})-9(3^y)=2(3^y)^3-5(3^y)^2-9(3^y)$$

anthonyym · Answer

I don't think that's equivalent Bobo

anonymous · Answer

No it's not equivelant. (3y)^3 is 27y^3, not what the original is.

welshfella · Answer

i#m not sure what is wanted here.  Do you want  everything to have an exponent y?

bobo-i-bo · Answer

It is, since:
$$3^{3y}=3^{y+y+y}=3^y3^y3^y=(3^y)^3$$

anonymous · Answer

3^(y+y+y) not equal to 3^y * 3^y * 3^y

anthonyym · Answer

Yeah it's equivalent nevemind

anonymous · Answer

Well the question says to find the values of y for which that equation = 0

welshfella · Answer

$$for example 3^{3y}= 27^{y}$$

bobo-i-bo · Answer

It is always true that:
$$a^{b+c}=a^ba^c$$

anonymous · Answer

I did not know that.. Thank you.

bobo-i-bo · Answer

welshfella · Answer

well you can do that using bobo's first post

anonymous · Answer

Yeah I think I've figured it out.

bobo-i-bo · Answer

So the clever thing to do here, is to use the substitution $z=3^y$, and then you should try and solve for z...

welshfella · Answer

let 3^y = a
2a^3 - 5a^2 - 9a = 0


solve for a 

then put a = 3^y

anonymous · Answer

Yup, x was 3, 3/2, and -2, therefore y = 1, 1/2, and -2/3

welshfella · Answer

a (2a^2 - 5a - 9) = 0

anonymous · Answer

Wait no that's wrong it's 3^y not 3y. Thanks.

welshfella · Answer

how did you get those results for x?