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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)
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
I assume the question is NOT about ln(a*b) = ln(a) + ln(b) because that is true for all positive real a and b. Instead, you want a solution to ln(a) * b = ln(a) + ln(b) or, ln(a) * (b-1) = ln(b) ln(a) = ln(b)/(b-1) ln(a) = ln[b1/(b-1)] Exponentiating, a = b1/(b-1) For any real number b > 1, a given by the above equation will meet your requirements.
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)]
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Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2
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ln(ln)
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)
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
You can also write this as ln(6 times 4)
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)
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.
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.
The correct formula for exponential interpolation is: y =ya*(yb/ya)^[(x-xa)/(xb-xa)], xa<x<xb and also, x=xa*[ln(yb)-ln(y)]/[ln(yb)-ln(ya)]+xb*[ln(y)-ln(ya)]/[ln(yb)-ln(ya)], ya<y<yb