document.write( "Question 718222: The graph of a logistic growth function y= (c)/(1+ae^-rx) reaches its point of maximum growth, where y=c/2 . Show that the x coordinate of this point is x= ln a/r .\r
\n" ); document.write( "\n" ); document.write( "I saw that y equaled to things so I set those equal to each other. \r
\n" ); document.write( "\n" ); document.write( "C/(1 + ae^-rx) = C/2\r
\n" ); document.write( "\n" ); document.write( "1+ae^-rx = 2\r
\n" ); document.write( "\n" ); document.write( "ae^-rx = 1\r
\n" ); document.write( "\n" ); document.write( "ln (ae^-rx) = ln (1)\r
\n" ); document.write( "\n" ); document.write( "ln (ae^-rx) = 0 ???\r
\n" ); document.write( "\n" ); document.write( "I am not sure how to solve this question, can you also explain the process. Thanks! \n" ); document.write( "

Algebra.Com's Answer #440817 by stanbon(75887) $\"\"$ $\"About$

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The graph of a logistic growth function y= (c)/(1+ae^-rx) reaches its point of maximum growth, where y=c/2 . Show that the x coordinate of this point is x= ln a/r .
\n" ); document.write( "I saw that y equaled two things so I set those equal to each other.
\n" ); document.write( "C/(1 + ae^-rx) = C/2
\n" ); document.write( "1+a*e^(-rx) = 2
\n" ); document.write( "ae^-rx = 1
\n" ); document.write( "e^(-rx) = 1/a
\n" ); document.write( "----
\n" ); document.write( "-rx = -lna
\n" ); document.write( "------
\n" ); document.write( "x = (1/r)ln(a)
\n" ); document.write( "OR
\n" ); document.write( "x = ln[a^(1/r)]
\n" ); document.write( "==================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "================== \n" ); document.write( "

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