Question

In a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is \(n = f\left( t \right) = {\bf{100}} \cdot {{\bf{2}}^{\frac{t}{{\bf{3}}}}}\).

(a) Find the inverse of the function and explain its meaning?

(b) When will the population reach 50,000?

Short answer

(a) The inverse function is \({f^{ - 1}}\left( n \right) = 3{\log _2}\left( {\frac{n}{{100}}} \right)\).

(b) The population will reach 50,000 at 26.9 hours.

Step-by-step solution

Change of Base Formula

Let b be any positive number, then the formula is:

\({\log _b}x = \frac{{\ln x}}{{\ln b}}\)

Determine the inverse of the function

Solve the equation for \(t\) as shown below:

\(\begin{aligned}n &= 100 \cdot {2^{\frac{t}{3}}}\\\frac{n}{{100}} &= {2^{\frac{t}{3}}}\\{\log _2}\left( {\frac{n}{{100}}} \right) &= \frac{t}{3}\\t &= 3{\log _2}\left( {\frac{n}{{100}}} \right)\end{aligned}\)

Thus, the inverse function is \({f^{ - 1}}\left( n \right) = 3{\log _2}\left( {\frac{n}{{100}}} \right)\).

Use the change of Base Formula to write the equation as shown below:

\(\begin{aligned}t &= {f^{ - 1}}\left( n \right)\\ &= 3 \cdot \frac{{\ln \left( {\frac{n}{{100}}} \right)}}{{\ln 2}}\end{aligned}\)

Thus, the function can determine how long it will take to obtain \(n\) bacteria.

Determine when the population will reach 50,000 

b)

The population will reach \(n = 50,000\) as shown below:

\(\begin{aligned}t &= {f^{ - 1}}\left( {50,000} \right)\\ &= 3 \cdot \frac{{\ln \left( {\frac{{50,000}}{{100}}} \right)}}{{\ln 2}}\\ &= 3 \cdot \ln \left( {\frac{{500}}{2}} \right)\\ &= 3 \cdot 8.9657\\ \approx 26.9\,\,{\mathop{\rm hours}\nolimits} \end{aligned}\)

Thus, the population will reach 50,000 at 26.9 hours.