ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x.
so we have ln y = x. Relace ln with log base e, and you should get y = ex
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e-x = 6Take the natural log of both sides:ln(e-x) = ln(6)-x = ln(6)x = -ln(6)So x = -ln(6), which is about -1.792.
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4
p=4-lnx Are you trying to solve for x? If so, then lnx=4-p so x=e^(4-p) Perhaps you left out some information?
ln(4x)=13 4x=e^13 x=(e^13)/4