Question Number 5580 by Rasheed Soomro last updated on 21/May/16
$$\mathrm{Given}\:\mathrm{that}\:{p}^{{n}} =\mathrm{16}{p},\:\mathrm{express}\:\mathrm{log}_{\mathrm{2}} {p}\: \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}\:. \\ $$
Answered by FilupSmith last updated on 21/May/16
$${p}^{{n}} =\mathrm{16}{p} \\ $$$${p}^{{n}−\mathrm{1}} =\mathrm{16} \\ $$$$\left({n}−\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} {p}=\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{4}} \right) \\ $$$$\mathrm{log}_{\mathrm{2}} {p}=\frac{\mathrm{4}}{{n}−\mathrm{1}} \\ $$