Can I rewrite this solution to a log growth question with less steps and text? I am working on a textbook exercise:
Given the log growth formula of $\frac{1000}{1+9e^{-0.6t}}$ find how long it takes for the population to reach 900.
The solution provided is 7.3 and I was able to arrive at this. But, reading over my solution I 'feel' like it's got too many steps and text and I just wondered if there was a simpler route to get my answer.
Here's my working:
$$900=\frac{1000}{1+9e^{-0.6t}}$$
$$1000=(900)(1+9e^{-0.6t})$$
$$\frac{1000}{900}-1=9e^{-0.6t}$$
(here is where I think I'm doing thing's 'wrong'. I have a fraction of a fraction on the left side, is there a better way to denote this? Can I somehow use a negative exponent instead of dividing by 9 for example? Or anything else? Or is what I have sound?)
$$\frac{\frac{1000}{900}-1}{9}=e^{-0.6t}$$
$$ln(\frac{1000}{900}-1)-ln9=-0.6t$$
(Again, here I feel like this expression is too cluttered, is there a simpler, less text way to express it?)
$$\frac{ln(\frac{1000}{900}-1)-ln9}{-0.6}=t$$
$$t=7.3$$
More generally, I'm wondering if I'm missing any opportunities to simplify my fraction expressions, either using rules of logs that I've missed or any other means?

You can indeed save some work.
Write
$$
\frac{1000}{900} -1 = \frac{10}{9} -1 = \frac{1}{9}
$$
as soon as it appears.
Then you reach
$$
\frac{1}{81} = e^{-0.6t}
$$
with just one natural logarithm to calculate.
