Solving the equation $a ^ b + b ^ a = 200$ Find $a$ and $b$,
$a ^ b + b ^ a = 200$
One of the  answers is $a = 1$ and $b = 199$.
Lets say $a, b$ belongs to $\mathbb{R}$ then there will be many solutions, for each $a$ there exist $b$, in $\mathbb{R}$. 
But how to find out $b$ for each $a$? I tried many ways, but not able to solve. Here is one of the approach I tried,
Try-1
$a ^ b + b ^ a = 200$
Take log on both sides
$b \log a + a \log b = \log 200$
After this I am not able solve (means all $b$'s are one side, and all $a$'s are another side)
Try-2
$a ^ b + b ^ a = 200$
(Trying for a special case, say $a = b$)
$a^a+a^a=200$
$2a^a = 200$
$a ^ a = 100$
Take log base 10 on both sides
$a \log a = \log 100$
$a \log a = 2$
stuck here, and not able to solve and not able to find $b$.
 A: This problem is a very hard one, and I don't think you can solve it with the elementary functions. The problem is that logarithms of sums don't play nice.
Also, you made this error in try 1
$$\log(a^b+b^a)\ne b\log(a)+a\log(b)$$
For your special case $a=b$, there's a solution which was noted by gammatester,
and it is
$$a=\frac{\ln100}{W(\ln100)}\approx3.597$$
I solved the pairs numerically using newtons method
There's a function which turns one of the pair into another, it's plotted below
The function is going to be its own inverse


This function looks like $\frac1x$, it even seems to have the same limits, lets compare those

What happens if change the $1$ to something else so it's equal at one place? I know the function has the solution $a=1$ $b=199$ so lets try $\frac{199}{x}$.

While that did cross our function at 199, it didn't really act like the function..
So I tried some other things like logarithms, it looks a bit more like the function, but I'm giving up now.
Also here's a mathematically impractical, but useful for approximations solution to your problem.
Define the sequence below:
$$a_0=1$$
$$a_{n+1}=a_n-\frac{a_n (\lambda^{a_n}+a_n^\lambda)}{a_n \lambda^{a_n} log(\lambda)+\lambda a_n^\lambda}$$
Then the function which transforms one from a working pair $(a,b)$ into another
$$f(\lambda)=\lim_{n\rightarrow\infty}a_n$$
