Fix get_numpy_tensor for paddle.Tensor.__rpow__ #519
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设MAX是dtype支持的最大值，B是常数底，不妨设B>1
则
y = B^x
dy/dx = y*lnB
为了保证不溢出，需要y<MAX，dy/dx <MAX
令y = MAX，则
B^x=MAX
xlnB=lnMAX
x = lnMAX/lnB
令dy/dx = MAX
则 x = ln(MAX/lnB)/lnB
注：
当0<B<1时，可以通过1/B限制，因为此时y和dy/dx取得最大值的时机是x取最小值，x的最小值和最大值互为相反数，所以这与令B变为倒数是等价的