Last Activity: 13 Years ago
ln(ye^x)=3
ye^x=e^3 (as per the property, logea=b means a=eb )
now, y=e^(3-x)
dy/dx=e^(3-x)d(3-x)/dx (using chain rule)
:. dy/dx=-e^(3-x)
hope this helps....;)
ln(yex)=3
=> lny + lnex =3 (using property)
=> lny + x*lne =3
=> lny + x =3 (lne=1; using property-same baseand no.)
now diff;
dy/dx*1/y +1 =0
so dy/dx= -y
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