The Solution:
From the given graph, we have that
[tex]\begin{gathered} \text{ When t=0, p=}2 \\ \text{ When t=0.1, p=1} \end{gathered}[/tex]
We need to find the logarithmic function that describes the situation.
[tex]\begin{gathered} p=\log _at \\ \text{ Where} \\ p=PH\text{ function} \\ t=\text{amount of hydronium ions} \end{gathered}[/tex]
So, the logarithmic function is
[tex]p=\log _at[/tex]
To find the value of a, we shall use the initial values:
[tex]\begin{gathered} p=1\text{ when t=0.1} \\ \text{Substituting, we get} \\ 1=\log _a0.1 \\ a^1=0.1 \\ a=0.1 \end{gathered}[/tex]
Thus, the logarithmic function is
[tex]p=\log _{0.1}t[/tex]
Writing the logarithmic function as an exponential function, we get
[tex]t=0.1^p[/tex]
The pool maintenance man wants to raise the amount of hydronium ions to 0.50 , that is, t=0.50, find the PH level, that is, the value of p.
[tex]\begin{gathered} \text{ When t=0.50},\text{ find p} \\ \text{Substituting into the function, we get} \\ 0.50=0.1^p \\ \text{ Taking the logarithm of both sides, we get} \\ \log _{}0.50=\log _{}0.1^p \\ \log _{}0.50=p\log _{}0.1 \end{gathered}[/tex][tex]p=\frac{\log _{}0.50}{\log _{}0.1}=0.3010[/tex]
Therefore, the PH level will be 0.3010.