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I'm working through an introduction to differential equations and we saw this very basic equation (let's say $x$ is in dollars and $t$ is in years:

$\dot{x}(t) - kx(t)=-a$

Without running through all the details, I solved this and got

$x(t)=\frac{a}{k}+(x_0-\frac{a}{k})e^{kt}$

Okay, so far, I understand the units.

(1) $a$ is in dollars per year and $k$ is in 1/year so $a/k$ is in units of dollars.

(2) $x_0$ is in units of dollars so $(x_0-\frac{a}{k})$ is in units of dollars.

(3) $k$ is in units of 1/year and $t$ is in units of year so the exponential has no units.

My problem comes when I try to solve for $x(t)=0$

$0=\frac{a}{k}+(x_0-\frac{a}{k})e^{kt}$

$(\frac{a}{k}-x_0)e^{kt}=\frac{a}{k}$

$e^{kt}=\frac{a}{a-kx_0}$

$kt=ln(\frac{a}{a-kx_0})$

$t=\frac{1}{k}ln(\frac{a}{a-kx_0})$

Here, I'm still good - inside the log the units cancel out and $k$ is 1/years so $1/k$ is in units of years.

But, now for fun, I use the $ln$ rules and get:

$t=\frac{1}{k}[ln(a)-ln(a-kx_0)]$

And this is where I get confused. Inside each natural log, I get units of dollars/year - but how do you interpret the units?

roundsquare
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    A sort of "cheat" to work around is to rewrite $\frac{a}{a-kx_0} = \frac{a , $/yr}{a-kx_0 , $/yr} \times \frac{1 , $/yr}{1 , $/yr}$ and so when you use the log terms, it's $\ln(\frac{a , $/yr}{1 , $/yr} )$ for the first term and likewise for the second. I call it a "cheat" but it's actually required for this to work, you divide by a reference standard (here being $1/yr). https://math.stackexchange.com/questions/238390/units-of-a-log-of-a-physical-quantity here is a reference for more information. – Osama Ghani Mar 29 '20 at 00:21
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    (1) Osama Ghani's comment is correct. Only numbers have logarithms, not numbers-times-units. (2) In your example (and others like it), the actual units won't matter. If you change units by a factor $c$, both terms $\ln a$ and $\ln(a-kx_0)$ will increase by $\ln c$ and so their difference won't change. – Andreas Blass Mar 29 '20 at 02:37

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