J Johnwill New member Joined Sep 24, 2006 Messages 29 Nov 8, 2006 #1 hello, i need some assistance on this problem: y = e^(1+lnx) 1st. I brought down the (1+lnx) by using natural log on both sides. lny*y'=(1+lnx)*lne*1/x y'/y=(1/x)*1*-1/x^2 y'=e^(1+lnx)*-1/x^3 what do i do next?
hello, i need some assistance on this problem: y = e^(1+lnx) 1st. I brought down the (1+lnx) by using natural log on both sides. lny*y'=(1+lnx)*lne*1/x y'/y=(1/x)*1*-1/x^2 y'=e^(1+lnx)*-1/x^3 what do i do next?
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 8, 2006 #2 \(\displaystyle \L y = e^{\left( {1 + \ln (x)} \right)} = \left( e^1 \right)\left( {e^{\ln (x)} } \right) = \left( e \right)\left( x \right)\)
\(\displaystyle \L y = e^{\left( {1 + \ln (x)} \right)} = \left( e^1 \right)\left( {e^{\ln (x)} } \right) = \left( e \right)\left( x \right)\)
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 8, 2006 #4 Johnwill said: how did you get that e^lnx = x Click to expand... My goodness, you don't understand the relation of e<SUP>x</SUP> and ln(x). You need to read you textbook!
Johnwill said: how did you get that e^lnx = x Click to expand... My goodness, you don't understand the relation of e<SUP>x</SUP> and ln(x). You need to read you textbook!
J Johnwill New member Joined Sep 24, 2006 Messages 29 Nov 8, 2006 #5 nvm, gotcha Thanks for the warning?
