The statistical issues here are relatively minor once you take into account subject-matter knowledge. What you are mainly looking for are effects of experimental changes in medium components upon cell proliferation, and hints from changes in medium composition about what might be limiting proliferation.
I'd recommend focusing on intelligently chosen experimental manipulations, for which the outcome variable would be the cell proliferation rate or the number of cells after your 5 days. That is simple to evaluate with standard statistical methods, as there is one major outcome and only a few experimental manipulations. The Technical Perspective by the Pollards, Molecular Biology of the Cell 2019; 30: 1359-1368, should point you in the correct direction for those statistical analyses.
The changes in medium composition over time, however, can help design those experimental manipulations. Let's examine them in detail.
Let's say that you have a system with mammalian cells attached to beads suspended in culture medium, similar to Chen et al., Cytotherapy 2015; 17: 163-173. Based on data similar to what's in that paper and basic principles of cell biology, you can make some simplifying assumptions that allow for analysis of changes of different types of components in the culture medium over time. Let's assume that the cultures are humidified and gassed at constant oxygen and carbon dioxide pressures.
Volume
Even at cell confluence you are unlikely to have more than $10^6$ cells per milliliter of medium. If individual cells have volumes of $10^{-11}$ liter (10 picoliters), then the total volume within cells per milliliter of total fluid is 10 microliters, only about 1% of total volume. Within about 1%, we can ignore changes in extracellular volume as cells increase in number.
Cell number
Let $N(t)$ be the total number of cells per milliliter at time $t$. If the cells are in an exponential proliferation phase as you indicate, then comparing time $t_1$ against $t_0$ gives:
$$N(t_1)=N(t_0) 2^{(t_1-t_0)/\tau}, $$
where $\tau$ is the cell-number doubling time. Based on cell-number values from consecutive sampling, you can estimate the value of $\tau$:
$$\tau = \frac{t_1-t_0}{\log_2 N(t_1)-\log_2 N(t_0)}.$$
In practice, the cell-doubling times aren't always constant through 5 days. They tend to be longer at early times as cultures become established, shorten at intermediate times in true exponential proliferation phase, then lengthen as cells approach confluence or limits of nutrients.
Don't over-react to small variations in estimates of $\tau$ between time periods, however. Although the estimates of total numbers of cells per milliliter can be large, those estimates are typically based on actual counts of just a few dozens to hundreds of cells. The actual number of counts limits the precision of of those estimates. If counts are Poisson-distributed, then the variance in the number of counts $n$ equals the mean and the coefficient of variation (standard deviation divided by the mean) can be estimated as $1/\sqrt n$. An estimate of $N$ based on only 100 actual cell counts thus has an inherent 10% standard error.
Inorganic components
Sodium, potassium, calcium and chloride are taken up by cells from the culture medium but can't be converted to other forms. Intracellular sodium and chloride concentrations tend to be lower than extracellular, and total intracellular calcium (as opposed to free cytoplasmic calcium) is typically not too far from extracellular, so there shouldn't be much change in your sodium and calcium measurements.
Intracellular potassium concentration, however, is typically 30 to 50 times that of extracellular (about 150 mmol/liter intracellular, versus 3 to 5 extracellular). At confluence with a total cell volume of 10 microliters per milliliter of culture medium, the cells will contain 1.5 micromole of potassium, out of 3 to 5 micromole intially in the milliliter of medium. Depending on details of your system there could be substantial changes in extracellular medium potassium over time, particularly at late times.
For changes over time, assume that intracellular potassium concentration $K_c$ and average cell volume $v_c$ are constant over time. The extracellular potassium as a function of time, $K_e(t)$, is then:
$$K_e(t) = K_e(0) - K_c v_c N(t),$$
giving:
$$K_e(0) - K_e(t) = K_c v_c N(t).$$
As you are measuring $N(t)$ with an expected exponential increase, a log-log plot of the two sides of that last equation should give a reasonable estimate of intracellular potassium per cell, $K_c v_c$:
$$ \log(K_e(0) - K_e(t)) =\log K_c v_c + \log N(t). $$
The errors in these estimates might be large, particularly at early times. The variance in $\log K_c v_c$ is:
$$\text{Var} (\log K_c v_c) = \text{Var} (\log(K_e(0) - K_e(t))) + \text{Var}(\log N(t))$$
The variance of the log of a random variable can often be approximated by the ratio of the variance of the variable to the square of its mean; see this page. If $N(t)$ is based on $n_t$ counts, then the second term in that equation is approximately $1/n_t$; see above. If the variance in an estimate of $K_e$ is $v_K$, the first term is approximately:
$$\frac{2v_K}{(K_e(0) - K_e(t))^2}, $$
which can be pretty large if the change in $K_e$ is small relative to the measurement error. If you want to estimate $\log K_c v_c $ this way via a regression, it would be wise to do a variance-weighted regression.
You don't mention phosphate, which can be limiting. Inorganic phosphate from the medium is incorporated by cells into intracellular nucleotides and nucleic acids, phospholipids and phosphoproteins, leading to surprisingly high total cellular phosphorus equivalent to nearly 100 mmol/liter. An equation like that for potassium above could be informative, as extracellular phosphate concentration must be kept low to avoid precipitation of calcium phosphate.
Organic components
Organic components of the medium can be metabolized into different forms, leading to decreases or increases in extracellular concentrations over time.
Say that an organic metabolite $m$ in the medium has a conversion rate of $\gamma_m$ per unit time per cell (positive for production, negative for consumption). Then the instantaneous rate of conversion of $m$ is
$$\frac{dm}{dt}=\gamma_m N(t).$$
Integrating for the change in extracellular $m$ between time $t_0$ and $t_1$ gives:
$$m_e(t_1)-m_e(t_0) = \int_{t_0}^{t_1} \gamma_m N(t_0)2^{t/\tau} dt= \frac{\gamma_m \tau}{\log 2} N(t_0) 2^{(t_1-t_0)/\tau}= \frac{\gamma_m \tau N(t_1)}{\log 2} ,$$
where we assume a constant cell-number doubling time $\tau$ over that time interval. As you are already estimating the doubling time $\tau$ for that time interval, that equation provides an estimate of the conversion rate of metabolite $m$ per cell per unit time, $\gamma_m$. If metabolism per cell is in a steady state throughout your 5 days, that conversion rate should be relatively constant across your 5 time intervals.
The principles above show what limits the precision of a $\gamma_m$ estimate. The actual number of cell counts limits the precision of $N(t)$ and $\tau$. The error in measuring $m_e$ relative to the observed difference limits the precision of the left side of the equation.
Look carefully at those conversion rates over time for all of the organic medium components that you are measuring. Pay particular attention to components whose consumption per day per milliliter is a large fraction of the original amount per milliliter. Those components of the medium might become close to undetectable at later times unless their consumption drops as the cells turn to alternate sources. Those data should provide the most useful clues about what components might be limiting proliferation, which you then can investigate in defined experiments.
I don't think that much would be gained by trying to use "machine learning" tools on these extracellular concentration measurements. There are only a few dozen components you are evaluating; the values and plots should be informative on their own and guide you toward defined, definitive experimental manipulations with straightforward interpretations in terms of effects on cell proliferation.
Some examples
With rough numbers close to what Chen et al. reported, here's what you night find after 5 days without a medium change. Assume that $\tau$ is 1 day, a typical value for mammalian cells adapted to culture (one cell division per day), and is constant through the time period, and that you end up with 1 million cells per milliliter of medium. Then, with time in days, the change in extracellular concentration of metabolite $m$ based on the above equation is:
$$m_e(5)-m_e(0)= \frac{\gamma_m \times 1 \times 10^6}{.693}.$$
Glucose. Glucose conversion is about -5 micromoles per million cells per day. Extracellular glucose concentration would drop by over 7 micromoles/milliliter, or 7 millimol/liter. That's an appreciable fraction even of a high-glucose Dulbecco medium.
Lactate. Depending on the cell type and culture conditions, much of that glucose can be metabolized to lactate, with 2 lactate per glucose. At 7.5 micromoles lactate production per million cells per day, you could end up with over 10 millimol/liter lactate. The associated fixed (non-carbonic) acid production would acidify the medium to an extent depending on the medium's acid/base buffering capacity.
Glutamine. Chen et al. reported glutamine conversion of -2 micromoles per million cells per day, for a drop under the above assumptions of 3 millimol/liter in extracellular glutamine over 5 days. Much of that will show up as extracellular ammonium ion. Be aware that glutamine can be converted spontaneously to pyrrolidonecarboxylic acid and ammonium even without cells under culture conditions, so you might need to have a cell-free control to evaluate the cellular contribution to glutamine and ammonia changes.
Other amino acids. "Essential" amino acids can't be synthesized by mammalian cells. Follow them carefully. If your culture medium contains serum then this gets more complicated, as the cells in principle could hydrolyze the serum proteins to acquire those amino acids.
