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What is equivalent to ln 6 plus ln 4?
You can also write this as ln(6 times 4)
What is the derivative of x to the power of ln x?
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.
Integrate of ln x squared?
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
How do you differentiate exponential function?
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)
What interest rate would be necessary to double a 100 investment in 11 years?
Suppose the interest rate is r% then 2 = (1+r/100)11 so that ln(2) = 11*ln(1+r/100) Then ln(1+r/100) = ln(2) /11 so 1 +r/100 = exp[ln(2)/11] so r/100 = exp[ln(2)/11] - 1 r = 100*{exp[ln(2)/11] - 1} = 6.504 % approx.
How would you solve ln 4 plus 3 ln x equals 5 ln 2?
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
How do you work out Ln 24 - ln x equals ln 6?
18
Why is the symbol for natural log ln?
ln(ln)
If you have a number with the exponent x how do you find the answer?
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)
What is the derivative of y equals xlnx?
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
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
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)
What is equivalent to ln 6 plus ln 4?
You can also write this as ln(6 times 4)
How do you solve for x 3 ln x - ln 3x equals 0?
3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)
Simplify In e to the 7th power?
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.
Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 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.
What is the equation for exponential interpolation?
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
Do any positive real numbers alpha and beta exist such that lnalpha times beta equals lnalpha plus lnbeta and if so what are they?
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.
