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Question Number 16769 by tawa tawa last updated on 26/Jun/17

2n + p = w^a   make a the subject of the formular.

$$\mathrm{2n}\:+\:\mathrm{p}\:=\:\mathrm{w}^{\mathrm{a}} \\ $$$$\mathrm{make}\:\mathrm{a}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}. \\ $$

Commented by RasheedSoomro last updated on 26/Jun/17

2n + p = w^a   log(2n + p) = a log(w)  a=((log(2n + p))/(log w))

$$\mathrm{2n}\:+\:\mathrm{p}\:=\:\mathrm{w}^{\mathrm{a}} \\ $$$$\mathrm{log}\left(\mathrm{2n}\:+\:\mathrm{p}\right)\:=\:\mathrm{a}\:\mathrm{log}\left(\mathrm{w}\right) \\ $$$$\mathrm{a}=\frac{\mathrm{log}\left(\mathrm{2n}\:+\:\mathrm{p}\right)}{\mathrm{log}\:\mathrm{w}} \\ $$$$ \\ $$

Commented by tawa tawa last updated on 26/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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