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3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)
the natural log, ln, is the inverse of the exponential. so you can take the natural log of both sides of the equation and you get... ln(e^(x))=ln(.4634) ln(e^(x))=x because ln and e are inverses so we are left with x = ln(.4634) x = -0.769165
ln(x)+1. This can be found using the product rule: if f(x)=uv, f'(x)=u'v+v'u, so if u is x and v is ln(x), the product rule gives (1)(ln(x))+(x)(1/x)=ln(x)+1.
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------