# Math Help - Logarithmic Word Problem

1. ## Logarithmic Word Problem

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 · 2t/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:

$
\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}
$

$
\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....

$
\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!

2. Originally Posted by 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 · 2t/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:

$
\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}
$

$
\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....

$
\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!

Everythings you did up until the change of base formula looks good, I believe that the change of base formula is $\log_{b}{M}=\frac{\log{M}}{\log{b}}$

So your final solution should be $\frac{\log{\frac{t^3}{60^3}}}{\log2}=f^{-1}(n)$

Hope that helps.

3. Hello,

Originally Posted by 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 · 2t/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:

$
\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}
$

$
\begin{gathered}
3\cdot\log _2 \left( {\tfrac{n}
{{60}}} \right) = t \hfill \\
{\color{red}\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....

$
\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!
See the red part..

Why did you invert n and t ?? You can't do it... If you're asked $f^{-1}(n)$, just find t with respect to n, and that's what you had

$3\cdot\log _2 \left( {\tfrac{n}
{{60}}} \right) = t$

You should get :

$t=f^{-1}(n)=\frac{\log_{10} \frac{n^3}{60^3}}{\log_{10} 2}$

4. Thanks people!

For part b, would you set the original equation or the inverse equal to 70,000?

5. Originally Posted by RedBarchetta
Thanks people!

For part b, would you set the original equation or the inverse equal to 70,000?
Well, try to think a bit on it

n represents the population.

Here, you want the population to be 70000.

So n=70000

And you're looking for t.

Which formula gives t with respect to n ?

$t=\frac{\log_{10} \frac{n^3}{60^3}}{\log_{10} 2}$

Or

$n=60 \cdot 2^{\frac t3}$

?

6. $
\begin{gathered}
\frac{{\log _{10} \frac{{n^3 }}
{{60^3 }}}}
{{\log _{10} 2}} = 70,000 \hfill \\
\log _{10} \frac{{n^3 }}
{{60^3 }} = 70,000 \cdot \log _{10} 2 \hfill \\
\log _{10} \frac{{n^3 }}
{{60^3 }} = \log _{10} 2^{70,000} \hfill \\
\end{gathered}
$

$
\begin{gathered}
10^{\log _{10} \frac{{n^3 }}
{{60^3 }}} = 10^{\log _{10} 2^{70,000} } \hfill \\
\frac{{n^3 }}
{{60^3 }} = 2^{70,000} \hfill \\
\end{gathered}
$

$
\begin{gathered}
n^3 = 2^{70,000} \cdot 60^3 \hfill \\
n = \sqrt[3]{{2^{70,000} \cdot 60^3 }} \hfill \\
\end{gathered}
$

$
n = 60 \cdot 2^{\frac{{70000}}
{3}}
$

I can't even enter it into the calculator.

7. Remember what Moo said? Consider your OP:

Originally Posted by RedBarchetta
[tex]
3\cdot\log _2 \left( {\tfrac{n}
{{60}}} \right) = t \hfill \\[/Math]
That was fine. But then you tried to swap n for t:

$\log _2 \left( {\tfrac{{t^3 }}
{{60^3 }}} \right) = n$
That was not fine at all. You can't just switch variables like that. So put that out of your mind.

Going back:

Originally Posted by RedBarchetta
[tex]
3\cdot\log _2 \left( {\tfrac{n}
{{60}}} \right) = t \hfill \\[/Math]
Now, you need to get log or ln instead of $log_2$. So:

$\frac{3ln\frac{n}{60}}{ln 2}=t$

Or:

$\frac{3log\frac{n}{60}}{log 2}=t$

Either one will do. Now, plug in 70,000 for n and see what happens.