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Question Number 163881 by abdurehime last updated on 11/Jan/22

$${16}^{1-\mathrm{x}}{32}^{2\mathrm{x}+1}={128}^{2\mathrm{x}-1}\mathrm{solve}$$

Answered by Ar Brandon last updated on 11/Jan/22

$$2^{4(1-x)+5(2x+1)}=2^{7(2x-1)}$$$$8x=16,x=2$$

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