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Question Number 92247 by askask last updated on 05/May/20

How to convert the non−linear equations  to linear form?    y=(x/(c+mx))    y=ce^(mx)

$$\mathrm{How}\:\mathrm{to}\:\mathrm{convert}\:\mathrm{the}\:\mathrm{non}−\mathrm{linear}\:\mathrm{equation}{s} \\ $$$$\mathrm{to}\:\mathrm{linear}\:\mathrm{form}? \\ $$$$ \\ $$$${y}=\frac{{x}}{{c}+{mx}} \\ $$$$ \\ $$$${y}={ce}^{{mx}} \\ $$

Commented by Joel578 last updated on 06/May/20

Is this what you mean?    • y = c e^(mx)   ln y = mx + ln c  Y = ln y, M = m, X = x, C = ln c  ⇒ Y = MX + C    • y = (x/(c + mx))  (1/y) = ((c + mx)/x) = c (1/x) + m  Y = (1/y), C = c, X = (1/x), M = m  ⇒ Y = CX + M

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{what}\:\mathrm{you}\:\mathrm{mean}? \\ $$$$ \\ $$$$\bullet\:{y}\:=\:{c}\:{e}^{{mx}} \\ $$$$\mathrm{ln}\:{y}\:=\:{mx}\:+\:\mathrm{ln}\:{c} \\ $$$${Y}\:=\:\mathrm{ln}\:{y},\:{M}\:=\:{m},\:{X}\:=\:{x},\:{C}\:=\:\mathrm{ln}\:{c} \\ $$$$\Rightarrow\:{Y}\:=\:{MX}\:+\:{C} \\ $$$$ \\ $$$$\bullet\:{y}\:=\:\frac{{x}}{{c}\:+\:{mx}} \\ $$$$\frac{\mathrm{1}}{{y}}\:=\:\frac{{c}\:+\:{mx}}{{x}}\:=\:{c}\:\frac{\mathrm{1}}{{x}}\:+\:{m} \\ $$$${Y}\:=\:\frac{\mathrm{1}}{{y}},\:{C}\:=\:{c},\:{X}\:=\:\frac{\mathrm{1}}{{x}},\:{M}\:=\:{m} \\ $$$$\Rightarrow\:{Y}\:=\:{CX}\:+\:{M} \\ $$

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