another simultaneous log equation

so I stumbled upon this question
2log2(y)=log2(root3)+log2(x)
3^y=9^x
solve for x and y with the solutions being positive.
so ill show what I have done with this so far:
the first equation simplifies to
y^2=root3+x
the second equation simplfies to
log3(3^y) = log3(9^x)
ylog3(3)=xlog3(9)
y=2x
this is what i did next
y^2=root3+x
x = y^2 - root3
-------------------
y=2x
y=2(y^2 - root3)
y=2y^2-root3
2y^2 - y - root3 = 0
-------------------
I tried the quadratic and formula and factorising it but to no avail so I am assuming I did something wrong which I was hoping someone could point out? Thanks

so I stumbled upon this question
2log2(y)=log2(root3)+log2(x)
3^y=9^x
solve for x and y with the solutions being positive.
so ill show what I have done with this so far:
the first equation simplifies to
y^2=root3+x
the second equation simplfies to
log3(3^y) = log3(9^x)
ylog3(3)=xlog3(9)
y=2x
this is what i did next
y^2=root3+x
x = y^2 - root3
-------------------
y=2x
y=2(y^2 - root3)
y=2y^2-root3
2y^2 - y - root3 = 0
-------------------
I tried the quadratic and formula and factorising it but to no avail so I am assuming I did something wrong which I was hoping someone could point out? Thanks

Your simplification of the first equation is incorrect. You can't work with logs in that term-by-term fashion.

Recall that

$\log a + \log b = \log ab$