#### Mathmasteriw

##### Junior Member

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- Thread starter Mathmasteriw
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\(\displaystyle V= log_e(2t)\)

Let \(\displaystyle V=log_e(u)\) where \(\displaystyle u=2t\)

Therefore \(\displaystyle \frac{dV}{du}=\frac{1}{u}\) and \(\displaystyle \frac{du}{dt}=2\)

\(\displaystyle \frac{dV}{dt}=\frac{dV}{du}\frac{du}{dt} = \frac{1}{u} *2=\frac{2}{u} = \frac{2}{2t} = \frac{1}{t}\)

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O great ! Yes that looks a much better way to present the answer!

Thank you Harry!

Thank you Harry!

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V(t) is NOT a function of 'x' !Hi Guys and gals,

Is this correct?

Is there a better way to write it out?

Thanks

View attachment 24279

Why do you have 'dV/dx' in your work ?!!

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ALog

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Thanks Jomo!!

ALog_{e}u-->(A/u)du/dx