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Q: What is the range of ln x y-1?

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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

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 -----------------------------------------

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

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Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x

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For the function: y = x^x^x (the superscript notation on this text editor does not work with double superscripts) To solve for the derivative y', implicit differentiation is needed. First, the equation must be manipulated so there are no x's raised to x's on the right side of the equation. So, both sides of the equation must be input into a natural logarithm, wherein we can use the properties of logarithms to remove the superscripted powers of the right side: ln(y) = ln(x^x^x) ln(y) = xxln(x) ln(y)/ln(x) = xx ln(ln(y)/ln(x)) = xln(x) eln(ln(y)/ln(x)) = exln(x) ln(y)/ln(x) = exln(x) ln(y) = ln(x)exln(x) Now there are no functions raised to functions (x's raised to x's). Deriving this equation yields: (1/y)(y') = ln(x)exln(x)(x(1/x) + ln(x)) + exln(x)(1/x) = ln(x)exln(x)(1 + ln(x)) + exln(x)(1/x) = exln(x)(ln(x)(1+ln(x)) + (1/x)) Solving for y' yields: y' = y[exln(x)(ln2(x) + ln(x) + (1/x))] or y = xx^x ln(y) = ln(x)x^x ln(y) = xxln(x) ln(y) = exlnxln(x) y'/y = exlnx[ln(x) + 1)ln(x) + exlnx(1/x) y' = y[exlnx(ln2(x) + ln(x) + 1/x)] y' = xx^x[exlnx(ln2(x) + ln(x) + 1/x)]

{3x +y =1 {x+y= -3

I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)

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.

A logarithm is an exponent.Assume 1 ≠ a > 0 and x > 0Definition of Logarithmic Function with base a:y = logax ↔ ay = xln(ay) = ln x = y ln ay = ln x/ln aDefinite logax = ln x/ln aProperties:The domain of f is all positive real numbers.The range of f is all positive real numbers.f(1) = 0f is an increasing function when a > 1 and decreasing if 0 < a < 1. From the definition of logarithm, it's obvious thatf(x) = logax and f(x) = ax are inverse functions.

There are many calculations that could be done: =SUM(Y1:Y10) =AVERAGE(Y1:Y10) =MAX(Y1:Y10) =MIN(Y1:Y10) =COUNT(Y1:Y10)

int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]

2x = 5 then x*ln(2) = ln(5) so that x = ln(5)/ln(2) = 2.3219 approx.

How do you solve ln|tan(x)|=ln|sin(x)|-ln|cos(x)|? Well you start by........