If $a^b+b^a=800$, then find $a$ and $b$. I've solved many of this type of equations (without steps) by noticing certain pattern in the answers. 
Can someone please give the answer with proper steps.
P.S. I know the answer will be $799$ and $1$ but I'm asking for the steps.
 A: If $a,b\ge 2$, then $a,b\le 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2\le a\le b\le 10$, which give no solution.
A: For any equation of the form $a^b+b^a=k$ where $k \in \mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$
A: For any $a\in (0, 799]$ there exists a real number $b\in\mathbb{R}$ such that $a^b+b^a = 800$.  
To see why, let $f(b):=a^b+b^a$.  Now $f(1) = a + 1<800$.  Also, $f(b)>b^a$ and $\lim_{b\rightarrow\infty} b^a = +\infty$.  These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).  
(Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)


*

*If a=2, then b is approximately 9.47224.

*If a=3, then b is approximately 5.82613.

*If a=4, then b is approximately 4.37832.

*If a=5, then b is approximately 3.49605.

*If a=6, then b is approximately 2.91562.

*If a=7, then b is approximately 2.52955.

*If a=8, then b is approximately 2.2636.

*If a=9, then b is approximately 2.07237.

*If a=10, then b is approximately 1.92944.


All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.

