Why does the approximation for exponents $(a+b)^c \approx a^{c-bc} (a+1)^{cb}$ work? I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.
Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.
Break $x$ into two parts, an integer part, and an additive. For example $3.7\to 3 + 0.7$.
Therefor, $x\to(a+b)$ where $a$ is an integer part.
The approximation formula is:
$$(a+b)^c \approx a^{c-bc} (a+1)^{cb}$$
Or in my original form: 
$$(a+b)^c \approx ((a+b)-b)^{c(1-b)} ((a+b)+(1-b))^{cb}$$
It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.
For example:
$$37.5^{28} ≈ 37^{14}\cdot38^{14}$$
And sure enough, if we divide both parts, the ratio is 1.002 which is very close. 
Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.
Can anyone shed some light as to why this approximation method I've stumbled upon works?
 A: As a general rule, if you see a lot of products and exponents it may clear things up to take logs.  In your case, 
$$(a+b)^c ≈ a^{c-bc} \cdot (a+1)^{cb}$$
becomes
$$c \log(a+b) \approx  c(1-b) \log(a) + cb \log(a+1)$$
or just
$$\log(a+b) \approx (1-b) \log(a) + b \log(a+1).$$
This is equivalent to doing a linear interpolation of $\log x$ between the points $a$ and $a+1$.  This will be pretty accurate when $a$ is large because $\log x$ will be close to linear between $a$ and $a+1$.
A: Write your approximation
$$a^{c-bc} \cdot (a+1)^{cb}=a^c\left(1+\frac 1a\right)^{bc}$$
and your approximation is $$\left(1+\frac 1a\right)^{b}\approx 1+\frac ba$$
Which is the first two terms of the binomial expansion.  It will be reasonably accurate when $\frac ba \ll 1$  The next term is $\frac {b(b-1)}{2a^2}$
A: The answer is obvious.  If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^{c-1}+ ...$ for both forms of the equation.  Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant.  That is, for larger a, $a^n>>a^{n-1}$.
