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**Hint:**In this question we will consider the salt content at the time $t$ to be $u$$lbs$and then apply the rate of change using differential equations. We will consider the concentration of brine to be $c$at time $t$. And since the concentration of brine in the solution is $50$ times greater than the quantity of salt, we will substitute it in the equation and use integration and eliminate the constants and solve for the value of $\lambda $ which is the required solution.

**Complete step by step answer:**

Let the content of salt at time $t$ be $u$$lbs$ therefore, the rate of change is $\dfrac{du}{dt}$

And since the concentration is $2$ pounds per gallon, we have the quantity of salt as:

\[\Rightarrow \text{2 }gal\text{ }\!\!\times\!\!\text{ 2 }lbs\text{=4 }lbs/\min \]

If $c$ is the concentration of brine at time $t$, the rate at which the salt content decreases due to the out flow will be:

\[\Rightarrow \text{2 }gal\text{ }\!\!\times\!\!\text{ c }lbs\text{=2c }lbs/\min \]

So, the total rate of change will be the salt content minus the outflow therefore:

$\Rightarrow \dfrac{du}{dt}=4-2c$

And since there is no increase in volume of the liquid concentration the quantity of salt is $50$ times the salt content therefore, $c=\dfrac{u}{50}$

On substituting it in the expression, we get:

$\Rightarrow \dfrac{du}{dt}=4-\dfrac{2u}{50}$

Now on separating the variables in the equation, we get:

$\Rightarrow dt=\dfrac{25du}{100-u}$

Now on integrating, both the sides, we get:

$\Rightarrow \int{dt=25\int{\dfrac{du}{100-u}}}$

On integrating using the formula $\int{\dfrac{1}{x}=\log x+k}$, we get:

\[\Rightarrow t=-25\ln (100-u)+K\to (1)\]

now initially when time is $0$ the salt content is $0$ therefore, on substituting, we get:

\[\Rightarrow 0=-25\ln (100-0)+K\]

On simplifying, we get:

\[\Rightarrow K=25\ln (100)\]

On substituting the value of $K$ in equation $(1)$, we get:

\[\Rightarrow t=-25\ln (100-u)+25\ln (100)\to (1)\]

On rearranging the terms, we get:

$\Rightarrow t=25\ln \left( \dfrac{100}{100-u} \right)$

Now the quantity of salt increases from $40$ to $80$ therefore, at time ${{t}_{1}}=25\ln \left( \dfrac{100}{60} \right)$ and at time ${{t}_{2}}=25\ln \left( \dfrac{100}{20} \right)$

Now the required time is:

$\Rightarrow {{t}_{2}}-{{t}_{1}}$

On substituting the values, we get:

$\Rightarrow 25\ln \left( \dfrac{100}{60} \right)-25\ln \left( \dfrac{100}{20} \right)$

On simplifying, we get:

$\Rightarrow 25\left( \ln 5-\ln \dfrac{5}{3} \right)$

On simplifying, we get:

$\Rightarrow 25\ln 3$

Now the value of $\ln 3=1.0986$therefore, on substituting, we get:

$\Rightarrow 25\times 1.0986$

On multiplying, we get:

$\Rightarrow \text{26 min 28 s}$

Now on converting the time into seconds, we get:

$\Rightarrow 1648s$

Now we know that the time taken is $206\lambda $

On substituting, we get:

$\Rightarrow 206\lambda =1648$

On simplifying, we get:

$\Rightarrow \lambda =8$, which is the required solution.

**Note:**In this question we are using the differential equal to find the difference in quantity of salt between two time periods therefore, we are finding the growth in the system since the quantity is increasing. If the quantity was decreasing, the condition is called decay. Whenever integration or differentiation is used, the log used in the natural log which is log to the base $e$.

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