B Why Does ln(y) = ln(bεax) Cause Issues in Derivation?

[Elias Waranoi]

I know that deriving y = bε^ax gives dy/dx = abε^ax, I also know that the inverse of dx/dy equals dy/dx, so how come I don't get dy/dx = abε^ax when I do the following:

ln(y) = ln(bε^ax) = ax ln(bε) -> x = ln(y)/(a ln(bε)) -> dx/dy = 1/(ay ln(bε)) -> dy/dx = ay ln(bε) = abε^ax ln(bε) ≠ abε^ax

What am I doing wrong?

 

[andrewkirk]

$\log y = \log \left(be^{ax}\right) = ax \log(be)$

This is where it goes wrong. It should be:
$$\log y = \log \left(be^{ax}\right) = \log b+ax$$

 

[Elias Waranoi]

This is where it goes wrong. It should be:
$$\log y = \log \left(be^{ax}\right) = \log b+ax$$

So like this: ln y = ln(bε^ax) = ln(b) + ax ln(ε) and because ln(ε) = 1 then ln y = ln(b) + ax. Thank you very much

 

[Mark44]

ln(y) = ln(bε^ax) = ax ln(bε)

So like this: ln y = ln(bε^ax) = ln(b) + ax ln(ε) and because ln(ε) = 1 then ln y = ln(b) + ax.

The simple answer, and to elaborate on what @andrewkirk said, what you are doing wrong in the first quote above is treating $be^{ax}$ as if it were $(be)^{ax}$ when you take the log. The expressions $be^{ax}$ and $(be)^{ax}$ are different, just as $2x^3$ and $(2x)^3$ are different.

So $\ln(be^{ax}) = \ln(b) + \ln(e^{ax}) = \ln(b) + ax$.

BTW, you should use the letter 'e' for the exponential base, not ε ("epsilon"). At least that's what I think you mean.