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16-1-x-32-2x-1-128-2x-1-solve-




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

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