... and without logarithms or knowing any powers of 2 other than the most trivial one ...

\(\displaystyle{2}^{{{4}{x}+{1}}}={128}\)

\(\displaystyle{2}^{{{4}{x}+{1}-{1}}}={2}^{{{4}{x}}}={\frac{{{128}}}{{{2}}}}={64}\)

\(\displaystyle{2}^{{{4}{x}-{1}}}={32}\)

\(\displaystyle{2}^{{{4}{x}-{2}}}={16}\)

\(\displaystyle{2}^{{{4}{x}-{3}}}={8}\)

\(\displaystyle{2}^{{{4}{x}-{4}}}={4}\)

\(\displaystyle{2}^{{{4}{x}-{5}}}={2}\)

\(\displaystyle{2}^{{{4}{x}-{6}}}={1}={2}^{{0}}\)

so \(\displaystyle{4}{x}−{6}={0}\) and \(\displaystyle{x}={\frac{{{6}}}{{{4}}}}={\frac{{{3}}}{{{2}}}}\).

\(\displaystyle{2}^{{{4}{x}+{1}}}={128}\)

\(\displaystyle{2}^{{{4}{x}+{1}-{1}}}={2}^{{{4}{x}}}={\frac{{{128}}}{{{2}}}}={64}\)

\(\displaystyle{2}^{{{4}{x}-{1}}}={32}\)

\(\displaystyle{2}^{{{4}{x}-{2}}}={16}\)

\(\displaystyle{2}^{{{4}{x}-{3}}}={8}\)

\(\displaystyle{2}^{{{4}{x}-{4}}}={4}\)

\(\displaystyle{2}^{{{4}{x}-{5}}}={2}\)

\(\displaystyle{2}^{{{4}{x}-{6}}}={1}={2}^{{0}}\)

so \(\displaystyle{4}{x}−{6}={0}\) and \(\displaystyle{x}={\frac{{{6}}}{{{4}}}}={\frac{{{3}}}{{{2}}}}\).