Given that 3^x = 9^(y-1), how would I show that x = 2y-2?

3^x = 9^(y - 1)
3^x = (3^2)^(y - 1)
3^x = 3(2 x (y - 1))
3^x = 3^(2y - 2)
Take away the base of 3
x = 2y - 2

3^x = 9^(y - 1)
3^x = 3^[2(y - 1)]
x = 2(y - 1)
x = 2y - 2

3^x = (3²)^(y - 1)
3^x = 3^(2y - 2)
x = 2y - 2

You can use ln or exponent;
3^x=9^(y-1)
xln3=(y-1)ln9
xln3=(y-1)ln(3^2)
x=2(y-1)
x=2y-2

I would try plugging in a value for X into equation two, getting the y value and then seeing if that matches up in equation one.

like this...
0 = 2y - 2 simplifies to y =1
plugging into other equation gives...
3^0 = 9^(1-1) which of course simplifies to 1 = 1

Since that equation is true, the second equation is true if the first one is. Although, to prove it beyond a doubt do the next method.

SECOND METHOD:

You could also graph them and compare graphs

Third method:

You could also try to solve the first equation for the second.

3^x = 9^(y - 1)
3^x = 3^[2(y - 1)]
x = 2(y - 1)
x = 2y - 2

3^x = 9^(y-1),
3^x = 3^2(y-1),
Equate exponents and u get x = 2(y-1)
or x = 2 y- 2

You shoot yourself. I know I would.