Another way to do problem #7:

Regrouping the terms yields
8^x + 18^x -2*27^x = [2^(3x)-3^(3x)] + [2^x3^(2x)-3^(3x)]=(2^x-3^x)(2^(2x)+2^x3^x+3^(2x))+3^(2x)(2^x-3^x)=(2^x-3^x)(4^x+6^x+2*9^x) > 0. Since 4^x + 6^x + 2*9^x > 0 for all real x, we have 2^x > 3^x or (2/3)^x > 1 or x < 0.