Getting to grips with simple logarithmic equations Please, help me to understand my mistakes, and the logic, so I can once and for all understand and remember nuances - they do seem to slip away as the time passes; I thought this topic is clear to me, but, well, I see it's not. There are two questions. 
(1) Here is the example I found on the web:

$\$100$ go to $\$150$ in $5$ years

To see how, they suggest to:

$\dfrac{\ln\left(\frac{150}{100}\right)}{5} = 8.1\%$

When I try to see how they got here and why their approach "works", I get stuck - I don't see how they came to the natural logarithm (base epsilon) and how their math came about. 
I assume they started from something like this: 

$100 \cdot Y^5 = 150$ (although usually it would be $100 \cdot (1 + x\%)^5 = 150$)
$Y^5 = \dfrac{150}{100}$
$\log_Y 1.5 = 5$ (here I assume it can be $\ln$ or $\log$ - it shouldn't matter as far as I know (which is very little) about both)

I don't see how it should evolve.
Then I decided to use simple math to see if I get the same $\%$ without using logarithm.

$100 \cdot (1 + x)^5 = 150$
$(1 + x)^5 = 1.5$
$x = 1.5^{1/5} - 1 = 8.4\%$
which is more or less close to $8.1\%$

(2) I decided to use another path, and decided to use their result $8.1\%$ to see if I get $5$ years to get from $100$ to $150$ with the given percent. So I did the following:

$100 \cdot (1 + 0.081)^N = 150$

Then, using logarithm base $10$, I indeed get almost $5$ years:

$(1 + 0.081)^N = 1.5$
$\log_{1.081} 1.5 = 5.2$

 A: Their solution is incorrect. When you solve for $Y$ you get along the way that $$\log Y = \frac{\log \left( \frac{150}{100} \right)}{5}$$ which is approximately $0.81$ if you use the logarithm base $e$. 
Now, if $Y$ is close to $Y$ then $\log Y \approx Y-1$, so you get $Y \approx 1.081$. This isn't very accurate, though. The precise answer is obtained through exponentiation giving you
$$Y = e^{\log Y} = e^{\log \left( \frac{150}{100} \right) /5} = \left( \frac{150}{100} \right)^{1/5} \approx 1.084$$ as you determined.
A: If 100Y^5= 150 then Y^5= 1.5 and, taking the logarithm of both sides, base Y, $5= \log_Y(1.5)$.  Now you say "here I assume it can be ln or log".  NO, it can't. That is only true if you are using the logarithm "base Y" but that does not help because you do not know what Y is!  Instead, use the fact that, for a logarithm to any base, b, $\log_b(Y^5)= 5 \log_b(Y)$.  So for any base, $5 \log(Y)= \log(1.5)$ and $\log(Y)= \frac{\log(1.5)}{5}$.  
Of course, if the logarithm is base 10, that gives $Y= 10^{\frac{\log(1.5)}{5}}= (10^{\log(1.5)})^{1/5}= (1.5)^{1/5}$ and, if it is the "natural" logarithm, $Y= e^{\frac{\ln(1.5)}{5}}= (e^{\ln(1.5)})^{1/5}= (1.5)^{1/5}$.  So using logarithms here is just a "fancy", and overly complicated, way of taking the fifth root.
The calculator that comes with "Windows" gives $(1.5)^{1/5}= 1.08447177119769861374560992241$ which.
A: Their answer is incomplete, this answer shows how to complete it.
Starting with $100*(1+x)^5 = 150$, we can transform this to $$(1+x)^5 = \frac{150}{100}$$ and proceed as you did.  We could also go along the (corrected) lines of their approach take the natural logarithm (or any other) on both sides and arrive at $$5\ln(1+x)=\ln(1.5),$$ which leads to $$\ln(1+x)=\frac{\ln(1.5)}{5}$$ (which amounts to the 'log-returns' of $8.1\%,$ however we are not done yet. Taking $\exp$ on both sides, we get
$$x=\exp(\ln(1.5)/5)-1=0.0844=8.44\%$$ which coincides with your solution.
