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Math Help - Logarithmic Word Problem

  1. #1
    Member RedBarchetta's Avatar
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    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:


    <br />
\begin{gathered}<br />
  n = 60\cdot2^{t/3}  \hfill \\<br />
  \frac{n}<br />
{{60}} = 2^{t/3}  \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \log _2 \left( {2^{t/3} } \right) \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \frac{t}<br />
{3} \hfill \\ <br />
\end{gathered} <br />

    <br />
\begin{gathered}<br />
  3\cdot\log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = t \hfill \\<br />
  \log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n = f^{ - 1} (n) \hfill \\ <br />
\end{gathered} <br />

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

    <br />
\begin{gathered}<br />
  3 \cdot \log _2 \left( {\tfrac{t}<br />
{{60}}} \right) = n = f^{ - 1} (n) \hfill \\<br />
  \log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n \hfill \\<br />
  \frac{{\log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right)}}<br />
{{\log _2 \left( {10} \right)}} = \log _{10} \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) \hfill \\ <br />
\end{gathered} <br />

    Apparently I went wrong somewhere....

    Thank you!
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  2. #2
    Jen
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    Quote Originally Posted by RedBarchetta View Post
    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:


    <br />
\begin{gathered}<br />
  n = 60\cdot2^{t/3}  \hfill \\<br />
  \frac{n}<br />
{{60}} = 2^{t/3}  \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \log _2 \left( {2^{t/3} } \right) \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \frac{t}<br />
{3} \hfill \\ <br />
\end{gathered} <br />

    <br />
\begin{gathered}<br />
  3\cdot\log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = t \hfill \\<br />
  \log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n = f^{ - 1} (n) \hfill \\ <br />
\end{gathered} <br />

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

    <br />
\begin{gathered}<br />
  3 \cdot \log _2 \left( {\tfrac{t}<br />
{{60}}} \right) = n = f^{ - 1} (n) \hfill \\<br />
  \log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n \hfill \\<br />
  \frac{{\log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right)}}<br />
{{\log _2 \left( {10} \right)}} = \log _{10} \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) \hfill \\ <br />
\end{gathered} <br />

    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.
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by RedBarchetta View Post
    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:


    <br />
\begin{gathered}<br />
  n = 60\cdot2^{t/3}  \hfill \\<br />
  \frac{n}<br />
{{60}} = 2^{t/3}  \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \log _2 \left( {2^{t/3} } \right) \hfill \\<br />
  \log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = \frac{t}<br />
{3} \hfill \\ <br />
\end{gathered} <br />

    <br />
\begin{gathered}<br />
  3\cdot\log _2 \left( {\tfrac{n}<br />
{{60}}} \right) = t \hfill \\<br />
  {\color{red}\log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n = f^{ - 1} (n)} \hfill \\ <br />
\end{gathered} <br />

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

    <br />
\begin{gathered}<br />
  3 \cdot \log _2 \left( {\tfrac{t}<br />
{{60}}} \right) = n = f^{ - 1} (n) \hfill \\<br />
  \log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) = n \hfill \\<br />
  \frac{{\log _2 \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right)}}<br />
{{\log _2 \left( {10} \right)}} = \log _{10} \left( {\tfrac{{t^3 }}<br />
{{60^3 }}} \right) \hfill \\ <br />
\end{gathered} <br />

    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}<br />
{{60}}} \right) = t


    You should get :

    t=f^{-1}(n)=\frac{\log_{10} \frac{n^3}{60^3}}{\log_{10} 2}
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  4. #4
    Member RedBarchetta's Avatar
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    Thanks people!

    For part b, would you set the original equation or the inverse equal to 70,000?
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  5. #5
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    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}

    ?
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  6. #6
    Member RedBarchetta's Avatar
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    <br />
\begin{gathered}<br />
  \frac{{\log _{10} \frac{{n^3 }}<br />
{{60^3 }}}}<br />
{{\log _{10} 2}} = 70,000 \hfill \\<br />
  \log _{10} \frac{{n^3 }}<br />
{{60^3 }} = 70,000 \cdot \log _{10} 2 \hfill \\<br />
  \log _{10} \frac{{n^3 }}<br />
{{60^3 }} = \log _{10} 2^{70,000}  \hfill \\ <br />
\end{gathered} <br />

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

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

    <br />
n = 60 \cdot 2^{\frac{{70000}}<br />
{3}} <br />

    I can't even enter it into the calculator.
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  7. #7
    Senior Member
    Joined
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    Remember what Moo said? Consider your OP:

    Quote Originally Posted by RedBarchetta View Post
    [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 }}<br />
{{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 View Post
    [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.
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