Question 1058803
{{{y=100*e^(kx)}}} a format for making an exponential growth model.


Doubling Time is 0.32 hours;
{{{100*e^(k*2)=200}}}
Can you calculate k?
This is not a complete solution but maybe you have enough to continue.



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asked for more than just the above ...


{{{e^(2k)=2}}}
{{{ln(e^(2k))=ln(2)}}}
{{{2k=ln(2)}}}
{{{k=ln(2)/2}}}
{{{k=0.3466}}}
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MODEL:  {{{highlight(y=100*e^(0.3466x))}}}


The exponential doubling model graphed:
The x-values of hours below 0 do not mean much.


 
{{{graph(300,300,-10,5,-10,500,100*e^(0.3466x))}}}



The inverse would be TIME in hours as a function of POPULATION SIZE.  If you do that, then y becomes time, and x becomes population size.  Now, y represents an logarithmic function instead of a exponential function.  The quick way to get to this is to take the logarithmic equation and switch the x and the y, and solve for y.
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{{{100*e^(ky)=x}}}
{{{ln(100)+ln(e^(ky))=ln(x)}}}
{{{ky=ln(x)-ln(100)}}}
{{{ky=ln(x/100)}}}
{{{y=ln(x/100)/k}}}
{{{highlight(y=ln(x/100)/0.3466)}}}