How to move from powers to simple logarithms? I'm following a book that briefly moves from
$$16000 \times 2^{\displaystyle \left (-\frac{x}{24} \right )} = 1600$$
to
$$x = \frac{24 (\log(2) + \log(5))}{\log(2)}$$
adding the comments that
$$\log(1600) = 6\log(2) + 2\log(5) \\
\log(16000) = 7\log(2) + 3\log(5)$$
What general principles are used to achieve this and how to spot possible ways to apply a similar operation in the future?
 A: In general:
\begin{align*}
\log a^b &= b \log a\\
\log (a \cdot b) &= \log a + \log b
\end{align*}
Specific to the above problem:
$$\log(1600) = \log(2^6 \cdot 5^2) = \log(2^6) + \log (5^2) = 6 \log 2 + 2 \log 5$$
$$\log(16000) = \log(2^7 \cdot 5^3) = 7 \log 2 + 3 \log 5$$
So,
\begin{align*}
16000 \times 2^{-x/24} &= 1600\\
\log 16000 + (-x/24) \log 2 &= \log 1600\\
(-x/24)\log 2 &= -\log 2 - \log 5\\
x &=24 \cdot \frac{\log 2 + \log 5}{\log 2}
\end{align*}
A: I'm not really sure why anyone would simplify that equation that way. Perhaps there is more context in the book.
Generally you take logs when you have something in an exponent you want to take down to "ground level", for example because you want to solve for it. If I encountered that equation, I would simplify it like this, using the two identities angryavian already provided:
\begin{align*}
16000 \times 2^{\displaystyle \left (-\frac{x}{24} \right )} &= 1600 \\
2^{\displaystyle \left (-\frac{x}{24} \right )} &= \frac{1}{10} \\
-\frac{x}{24} &= \log_2\frac{1}{10} \\
-\frac{x}{24} &= -\log_2 10 \\
\frac{x}{24} &= \log_2 10 \\
x &= 24\log_2 10 \\
\end{align*}
You can get this into the form of the book (though I don't so a good reason to unless it has some particular requirement) by continuing (using the identity $\frac{\log a}{\log b} = log_b a$)
\begin{align*}
x &= 24\log_2 10 \\
&= 24\frac{\log 10}{\log 2} \\
&= 24\frac{\log 2 + \log 5}{\log 2} \\
\end{align*}
A: First, lets $\log$ both sides
$$\log\left(16000\times 2^{\left(-\dfrac{x}{24}\right)}\right)=\log(1600)$$
The rule $\log(a\times b)=\log(a)+\log(b)$ will help here. Now we can simplify 
$\log\left(16000\times 2^{\left(-\dfrac{x}{24}\right)}\right)$.
$$\log(16000)+\log\left(2^{\left(-\dfrac{x}{24}\right)}\right)=\log(1600)$$
Now we use the rule $\log(a^x)=x\log(a)$ to simplify $\log\left(2^{\left(-\dfrac{x}{24}\right)}\right)$.
$$\log(16000)-\dfrac{x}{24}\log(2)=\log(1600)$$
$$-\dfrac{x}{24}\log(2)=\log(1600)-\log(16000)$$
We can simplify the right side by using the rule that $\log(x)-\log(y)=\log\left(\dfrac{x}{y}\right)$.
$$-\dfrac{x}{24}\log(2)=\log\left(\dfrac{1600}{16000}\right)$$
$$-\dfrac{x}{24}\log(2)=\log\left(\dfrac{1}{10}\right)$$
We will use $\log(x)-\log(y)=\log\left(\dfrac{x}{y}\right)$ again on the right hand side.
$$-\dfrac{x}{24}\log(2)=\log(1)-\log(10)$$
Remember, $\log(1)=0$
$$-\dfrac{x}{24}\log(2)=-\log(10)$$
$$\dfrac{x}{24}\log(2)=\log(10)$$
$$\dfrac{x}{24}=\dfrac{\log(10)}{\log(2)}$$
$$x=24\left(\dfrac{\log(10)}{\log(2)}\right)$$
$$x=24\left(\dfrac{\log(2\times 5)}{\log(2)}\right)$$
$$x=24\left(\dfrac{\log(2)+\log(5)}{\log(2)}\right)$$
$$\displaystyle \boxed{x=\dfrac{24(\log(2)+\log(5))}{\log(2)}}$$
I hope that this post helped you.
