It is quite often to use the log transformation on your data, if your data are always positive (e.g. the price of something) and their scales varies drastically.
A simple criterion of whether you should use log transformation is whether you want to use a linear or log scale for your x-axis when you are plotting the histogram of your data.
This is likely to make your ANN work better if your data indeed look that way, because of one reason: Remember the motivation of batch normalization - ANN likes to have a standard normal distribution. You can make your distribution zero-centered with unit variance, but that does not make your distribution into a normal distribution, but your distribution might look more like a normal distribution if you use log transformation. You can check whether this is true from the histogram or the Kurtosis of your distribution.
The easy answer is that standardization and normalization are linear transformations of the data and any line is determined by two distinct points on it. Since columns 3 and 4 are constructed to include the points $(\min(\text{data}), 0)$ and $(\max(\text{data}), 1)$ (namely, $(2,0)$ and $(95, 1)$), they must result from identical transformations.
The algebraically rigorous answer manipulates the formulas. Recall that standardization of data $\mathbf{x} = x_1, x_2, \ldots, x_n$ replaces each $x_i$ with
$$y_i = \frac{x_i - \bar x}{s_x}$$
where $\bar x$ usually is the (arithmetic) mean of the data and $s_x$ often is a standard deviation of the data. (Generally, $\bar x$ may be any estimate of a central value and $s_x$ may be any estimate of the spread, but such generality is usually not intended.)
On the other hand, normalization replaces each $x_i$ with
$$z_i = \frac{x_i - \min(\mathbf{x})}{\max(\mathbf x) - \min(\mathbf x)}.$$
But since $s_x \gt 0$, division by $s_x$ is an increasing function. Subtracting any constant like $\bar x$ similarly is an increasing function. Therefore the extremes of $\mathbf x$ correspond to the extremes of $\mathbf y$. Consequently, normalization of $\mathbf y = y_1, y_2, \ldots,y_n$ produces
$$\eqalign{
z_i^\prime &= \frac{y_i - \min(\mathbf{y})}{\max(\mathbf y) - \min(\mathbf y)} \\
&= \frac{\frac{x_i - \bar x}{s_x} - \min_j(\frac{x_j - \bar x}{s_x})}{\max_j(\frac{x_j - \bar x}{s_x}) - \min_j(\frac{x_j - \bar x}{s_x})} \\
&= \frac{{x_i - \bar x} - \min_j({x_j - \bar x})}{\max_j({x_j - \bar x}) - \min_j({x_j - \bar x})} \\
&= \frac{x_i - \min(\mathbf{x})}{\max(\mathbf x) - \min(\mathbf x)} \\
&= z_i.
}$$
Best Answer
That is equivalent to normalizing only $X$ since the standardization step does not change the min/max values. Besides, these transformations are not associated with normality assumption.