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e^(3lnx)=e^[ln(x^3)]=x^3

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Q: What is e to the power of 3 ln x?
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How do you Differentiate and Integrate y equals 3 to the power of x?

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).


How do you Simplify e to the 3lnx power?

"e^3lnx = (e^3) * (e^lnx) = (e^3) * x = xe^3" If you actually plugged in a constant for your variable you will see why this is wrong. Your thinking of e ^ (3 + ln(x)) ... The correct answer is e^(3ln(x)) = e^(ln(x^3)) = x^3


What is X when raised to the 3-5i power and the answer is 23-14i?

X = 1.31356+0.612045*iSteps to solve, take the natural log of both sides:ln(X^(3-5i)) = ln(23-14i).(3-5i)*ln(X) = ln(23-14i). Convert 23-14i to exponential form: A*e^(iΘ) {A = 26.926 and Θ = -0.54679 radians}(3-5i)*ln(X) = ln(A*e^(iΘ))= ln(A) + iΘ = ln(26.926) - 0.54679i.divide by (3-5i): ln(X) = (ln(A) + iΘ) / (3-5i) = (3.2931 - 0.54679i)/(3-5i)So we have ln(X) = 0.370978 + 0.436033i, then:e^(ln(X)) = e^(0.370978 + 0.436033i) --> X = 1.31356+0.612045*i


What is the derivative of e the the power ln x?

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


If e to the power x equals 0.4634 find x?

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


What is the antiderivative of e to the power of one divided by x?

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


What is the derivative of sin x to the e to the xth power?

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


Solve e raised to the power of negative x equals 6?

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.


What is the derivative of e to the power ln x squared?

e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.


-3 plus ln x equals 5?

-3 + ln(x) = 5Add 3 to each side:ln(x) = 8x = e(8) = 2,980.958 (rounded)


Which exponential equation is equivalent to 3 equals ln x?

3=lnx e^3=x


How do you solve 3 ln x - ln 2 equals 4?

3lnx - ln2=4 lnx^3 - ln2=4 ln(x^3/2)=4 (x^3)/2=e^4 x^3=2e^4 x=[2e^4]^(1/3)