2

^{s}=2^{3x-4}-
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2^{s}=2^{3x-4}

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logarithm 2x=2 3x-4 (variable x is introduced here)

2^s=2^(3x-4) and now variable S;

2^s=2^(3x-4) and now variable S;

This problem seems a bit unclear:

If you mean to solve:

2^{X }= 2^{(3X - 4)}, then this is straight forward(merely equate the exponents(x = 3x - 4; in which case x = 2)

please resend

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