I was trying to find a general solution to the equation $a^x=x+b$. First, I used a substitution:
$$u=a^x\Longleftrightarrow x=\log_au$$
Then, it went as follows:
$$u=\log_a u+b$$ $$-b=\log_a u-u$$ $$-b=\log_a u-\log_a a^u$$ $$-b=\log_a ua^{-u}$$ $$a^{-b}=ua^{-u}$$ $$-a^{-b}=-ua^{-u}$$ $$-u=W\big(\!-\!a^{-b}\,\big)$$ $$u=-W\big(\!-\!a^{-b}\,\big)$$ $$a^x=-W\big(\!-\!a^{-b}\,\big)$$
I finally got the solution of:
$$x={{\ln\Big(\!-W\big(\!-\!a^{-b}\,\big)\Big)}\over{\ln a}}$$
The problem:
From a paper on the Lambert W function, it says that a general solution to the same equation is:
$$x=-b-{{W\big(\!-\!a^{-b}\;\ln a\,\big)}\over{\ln a}}$$
which is quite different from what I got. I thought that maybe it is the same thing, but I wasn't able to manipulate my answer to get to the paper's answer.
So, the question is, how do I prove or disprove that:
$${{\ln\Big(\!-W\big(\!-\!a^{-b}\,\big)\Big)}\over{\ln a}}=-b-{{W\big(\!-\!a^{-b}\;\ln a\,\big)}\over{\ln a}}$$
EDIT: As Ahmed noted in the comments, I made an error in my work, I mistaked $-ua^{-u}$ for $-ue^{-u}$, and incorrectly took the Lambert W function of it, instead of changing the exp base first. When I corrected myself, I managed to arrive at the correct answer.