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
3=lnx e^3=x
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)
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)
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
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).
"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
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
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
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
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
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-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.
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 + ln(x) = 5Add 3 to each side:ln(x) = 8x = e(8) = 2,980.958 (rounded)
3=lnx e^3=x
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)