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**RedBarchetta** If a bacteria population starts with 60 bacteria and doubles every three hours, then the number of bacteria after *t* hours is *n* = *f*(*t*) = 60 **·** 2*t*/3.

(a) Find the inverse of this function.

*f* -1(*n*) = ___________

(b) When will the population reach 70,000? (Round the answer to the nearest tenth.)

Part A Attempt:

$\displaystyle

\begin{gathered}

n = 60\cdot2^{t/3} \hfill \\

\frac{n}

{{60}} = 2^{t/3} \hfill \\

\log _2 \left( {\tfrac{n}

{{60}}} \right) = \log _2 \left( {2^{t/3} } \right) \hfill \\

\log _2 \left( {\tfrac{n}

{{60}}} \right) = \frac{t}

{3} \hfill \\

\end{gathered}

$

$\displaystyle

\begin{gathered}

3\cdot\log _2 \left( {\tfrac{n}

{{60}}} \right) = t \hfill \\

\log _2 \left( {\tfrac{{t^3 }}

{{60^3 }}} \right) = n = f^{ - 1} (n) \hfill \\

\end{gathered}

$

Now I need to convert to base 10, since my online homework program online has inputs for natural log and log base 10. So....

$\displaystyle

\begin{gathered}

3 \cdot \log _2 \left( {\tfrac{t}

{{60}}} \right) = n = f^{ - 1} (n) \hfill \\

\log _2 \left( {\tfrac{{t^3 }}

{{60^3 }}} \right) = n \hfill \\

\frac{{\log _2 \left( {\tfrac{{t^3 }}

{{60^3 }}} \right)}}

{{\log _2 \left( {10} \right)}} = \log _{10} \left( {\tfrac{{t^3 }}

{{60^3 }}} \right) \hfill \\

\end{gathered}

$

Apparently I went wrong somewhere....

Thank you!