It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
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
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
e^(ln 3) = 3
To find ln 2.33, you need a calculator. It is the solution of the equation e^x = 2.33. ln 2.33 = 0.84586 (using a calculator)
ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2
Take the natural logarithm (ln) of both sides of the equation to cancel the exponent (e). For example, ify=Aexlog transform both sides and apply the rules of logarithms:ln(y)=ln(Aex)ln(y)=ln(A)+ln(ex)ln(y)=ln(A)+xrearrange in terms of x:x=ln(y)-ln(A), or more simplyx=ln(y/A)
It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
e^(3lnx)=e^[ln(x^3)]=x^3
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
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
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.
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace