Question Number 163881 by abdurehime last updated on 11/Jan/22
$$\mathrm{16}^{\mathrm{1}−\mathrm{x}} \mathrm{32}^{\mathrm{2x}+\mathrm{1}} =\mathrm{128}^{\mathrm{2x}−\mathrm{1}} \:\:\:\:\mathrm{solve} \\ $$
Answered by Ar Brandon last updated on 11/Jan/22
$$\mathrm{2}^{\mathrm{4}\left(\mathrm{1}−{x}\right)+\mathrm{5}\left(\mathrm{2}{x}+\mathrm{1}\right)} =\mathrm{2}^{\mathrm{7}\left(\mathrm{2}{x}−\mathrm{1}\right)} \\ $$$$\mathrm{8}{x}=\mathrm{16}\:,\:{x}=\mathrm{2} \\ $$