Can anyone help me with this question:

( [(9^{2x-1}) (3^{3x})^2] / [(27^{x+2})^4] ) = 81^3

I know that x=-19 but I keep getting -1.5 and 8. Step by step would be really appreciated. Thank you.

\frac{(9^{2x-1}) (3^{3x})^2}{(27^{x+2})^4} = 81^3

Okay, put everything to base 3.

\frac{((3^2)^{2x-1}) (3^{3x})^2}{((3^3)^{x+2})^4} = (3^4)^3

Now, work out the powers:

\frac{(3^{4x-2}) (3^{6x})}{(3^{12x+24})} = (3^{12})

Now, you know that (a^x)(a^y) = a^{x+y} and \frac{a^x}{a^y} = a^{x-y}

Apply this:

3^{4x-2 + 6x - 12x - 24} = 3^{12}

Now, simplify:

3^{-2x-26} = 3^{12}

Equate the powers:

-2x - 26 = 12

This gives you:

x = \frac{12 + 26}{-2} = -19

Thanks once again! I now see where I was going wrong. After I worked out the powers, instead of adding the top powers, I was trying to multiply them. (Actually, I think I did try to add them at one point but still ended up with the wrong answer!) Oye! Lol!