Form a differential equation that solves the amount of salt at any time t as follows

\(\displaystyle\frac{{{d}{A}}}{{\left.{d}{t}\right.}}=\) rate in - Rate out

\(\displaystyle={2}\cdot{2.25}-\frac{{{A}{\left({t}\right)}}}{{70}}\cdot{2.25}\)

\(\displaystyle=\frac{4.5}{{{9}{A}{\left({t}\right)}}}/{280}\)

\(\displaystyle=\frac{{{1260}-{9}{A}}}{{280}}\)

\(\displaystyle\frac{{{d}{A}}}{{\left.{d}{t}\right.}}=\frac{{{1260}-{9}{A}}}{{280}}\), A(0)= 0

\(\displaystyle\frac{{{d}{A}}}{{{1260}-{9}{A}}}=\frac{{\left.{d}{t}\right.}}{{280}}\)

\(\displaystyle\int\frac{{{d}{A}}}{{{1260}-{9}{A}}}=\int\frac{{\left.{d}{t}\right.}}{{280}}\)

\(\displaystyle-\frac{1}{{9}} \ln{{\left({1260}-{9}{A}\right)}}=\frac{t}{{280}}+{C}\)

A(0)= 0

\(\displaystyle\Rightarrow{C}=-\frac{1}{{9}} \ln{{\left({1260}\right)}}\)

\(\displaystyle\Rightarrow \ln{{\left({1260}-{9}{A}\right)}}=-{9}\frac{t}{{280}}+ \ln{{\left({1260}\right)}}\)

\(\displaystyle\Rightarrow{1260}-{9}{A}={e}^{{-{9}\frac{t}{{280}}+ \ln{{\left({1260}\right)}}}}\)

\(\displaystyle\Rightarrow{1260}-{9}{A}={1260}{e}^{{-{9}\frac{t}{{280}}}}\)

\(\displaystyle\Rightarrow{140}-{A}={140}{e}^{{-{9}\frac{t}{{280}}}}\)

\(\displaystyle\Rightarrow{A}{\left({t}\right)}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)

Obtain the time at which the brine leaving the tank will contain 15 lbs of salt as follows

\(\displaystyle{A}{\left({t}\right)}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)

\(\displaystyle{15}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)

\(\displaystyle{140}{e}^{{-{9}\frac{t}{{280}}}}={140}-{15}\)

\(\displaystyle{e}^{{-{9}\frac{t}{{280}}}}=\frac{125}{{140}}\)

\(\displaystyle-{9}\frac{t}{{280}}= \ln{{\left(\frac{125}{{140}}\right)}}\)

\(\displaystyle{t}=-\frac{280}{{9}} \ln{{\left(\frac{125}{{140}}\right)}}\)

\(t= 3.53 min\)