Interpolation error and it's weird behaviour

Question

I am trying to find the value of $f(x)=\ln(1-3x)$ at $x=-1.5$ using Newton's interpolation method. I am given the points $-3$, $-2.4$, $-2.2$, $1.8$, $-0.5$, $0$, and using Newton's method gives $f(x)\approx P_{5}(x)=1.70843...$. Problem arises when I look into the error of this value, using the formula $$\left|f(x)-P_{5}(x)\right|\le \dfrac{M_{6}}{6!}\left|(x+3)(x+2.4)(x+2.2)(x+1.8)(x+0.5)x\right|$$ where $$M_6=\max_{-3\le x\le 0}\left|f^{(6)}(x)\right|$$. But, of course, $f^{(6)}(x)=\dfrac{ -3^6\cdot 5!}{(1-3x)^6}$, and maximum value of its absolute value is $3^6\cdot 5!=87480$, which gives that $$\left|f(-1.5)-P_{5}(-1.5)\right|\le52$$ So, in conclusion, the approximated value using interpolation gives amazingly good precision to the actual value, but error seems to be too big to believe it. My question is, first, did I do anything wrong, or have I misscalculated, and second, if I did everything correctly, would the substitution $3x=u$ solve this problem? Because it seems that the coefficient in front of $x$ is what's causing this weird behaviour, because it multiplies too much when using too many nodes for interpolation. Thanks in advance.