# Logarithmic Word Problem

• May 20th 2008, 10:04 AM
RedBarchetta
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!
• May 20th 2008, 10:12 AM
Jen
Quote:

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. :)
• May 20th 2008, 10:15 AM
Moo
Hello,

Quote:

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 (Wink)

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

You should get :

$t=f^{-1}(n)=\frac{\log_{10} \frac{n^3}{60^3}}{\log_{10} 2}$
• May 20th 2008, 12:08 PM
RedBarchetta
Thanks people!

For part b, would you set the original equation or the inverse equal to 70,000?
• May 20th 2008, 12:11 PM
Moo
Quote:

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 :p

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

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

? :D
• May 20th 2008, 01:19 PM
RedBarchetta
$
\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. (Headbang)
• May 20th 2008, 02:02 PM
hatsoff
Remember what Moo said? Consider your OP:

Quote:

Originally Posted by RedBarchetta
$$3\cdot\log _2 \left( {\tfrac{n} {{60}}} \right) = t \hfill \\$$

That was fine. But then you tried to swap n for t:

Quote:

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

That was not fine at all. You can't just switch variables like that. So put that out of your mind.

Going back:

Quote:

Originally Posted by RedBarchetta
$$3\cdot\log _2 \left( {\tfrac{n} {{60}}} \right) = t \hfill \\$$

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.