\(\frac {8^x-2^x}{6^x-3^x} = \frac {2^x(2^{2x}-1)}{3^x (2^x-1)}\)
= \(\frac {2^x (2^x-1)(2^x+1)}{3^x (2^x-1)}\)
= \(\frac {2^x (2^x+1)}{3^x}\)
\(\therefore\) \(\frac {2^x (2^x+1)}{3^x}\) = 2 (Given)
\(\Rightarrow\) \(2^{2x} +2^x = 2.3^x\)
For x = 0, \(2^{2x} + 2^x = 2^0 + 2^0 = 1+1 = 2\)
\(2.3^x = 2.3^0 = 2 \times 1 = 2\)
Hence, \(2^{2x}+2^x = 2.3^x \) then x = 0
