# B Why does this derivation fail?

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1. May 28, 2017

### Elias Waranoi

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

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

What am I doing wrong?

2. May 28, 2017

### andrewkirk

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

3. May 28, 2017

### Elias Waranoi

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

4. Jun 9, 2017

### Staff: Mentor

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.

Last edited: Jun 9, 2017