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If r and s are positive integers such that $(2^{r})(4^{s}) = 16$, then $2r +s =$
in Mathematics by Diamond (61,686 points) | 204 views

1 Answer

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(2^r).((2^(2s)) = (2^4)

2^(r + 2s) = (2^4)

r + 2s = 4

By trial and error of all the four positive integers between 1 and 4, we get:

2 + 2(1) = 4

Implying that:    r = 2;     s = 1

Therefore: 2r + s = (2x2) + 1 = 5
by Diamond (39,212 points)

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