Ik krijg wat anders:
\(
(5,0)\quad \rightarrow \quad a\cdot ^5\!\log(5+b) =0\\
(29,12)\quad \rightarrow \quad a\cdot ^5\!\log(29+b) =12\\\\
\frac{a\cdot ^5\log(5+b)}{a\cdot ^5\log(29+b)}=0=\frac{^5\!\log(5+b)}{^5\!\log(29+b)}\\\\
^5\!\log(5+b)=0\quad\rightarrow\quad 5+b=1\quad\rightarrow\quad b=-4\\\\
a\cdot ^5\!\log(29+b) =12 \quad\rightarrow\quad a\cdot^5\!\log(29+(-4)) =-12 \quad\rightarrow\quad a=-6\)