$\ln(x) -\ln(y) = kt$, make x the subject.

Question

So I thought you take the inverse function of the whole expression getting: $x - y = e^{kt}$ and so your final answer would be $x = e^{kt} + y$ but according to the answers in the book $x = ye^{kt}$. Where have I gone wrong?

And can you give a step-by-step method for working out the answer as I don't really understand how e relates to the ln() function.

Answer 1

You've incorrectly exponentiated the expression. $\ln x - \ln y = \ln ( \frac{x}{y})$. Now exponentiate both sides, and you should get the right answer.

Answer 2

HINT

Note that

$$\ln(x) - \ln(y) = kt\iff \ln(x) = \ln(y) + kt \iff e^{\ln(x)} = e^{\ln(y) + kt}=e^{\ln(y)}e^{kt}$$

or as an alternative

$$\ln(x) - \ln(y) = kt\iff e^{\ln(x) - \ln(y)} = e^{kt}\iff \frac{e^{\ln(x)}}{ e^{ln(y)}} = e^{kt}$$

Answer 3

$$x = e^{kt} + y \implies x -y = e^{kt} \implies \ln|x-y|=kt$$

Different from what you had in the begining...

$$\ln|x-y| \neq \ln|x|-\ln|y|$$