iillaa iillaa
  • 23-01-2016
  • Mathematics
contestada

Please show me how to do it.

Please show me how to do it class=

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Panoyin
Panoyin Panoyin
  • 23-01-2016
[tex]\frac{8^{2n - 3} * 2^{n - 3}}{4 * 2^{-n + 2}} = \frac{2^{3(2n - 3)} * 2^{n - 3}}{4 * 2^{-n + 2}} = \frac{2^{3(2n) - 3(3)} * 2^{n - 3}}{4 * 2^{-n + 2}} = \frac{2^{6n - 9} * 2^{n - 3}}{2^{2} * 2^{-n + 2}} = \frac{2^{(6n - 9) + (n + 3)}}{2^{2 - n + 2}} = \frac{2^{(6n + n) + (-9 + 3)}}{2^{-n + 2 + 2}} = \frac{2^{7x + (-12)}}{2^{-n + 4}} = \frac{2^{7x - 12}}{2^{-n + 4}} = 2^{8n - 16}[/tex]
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