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How do you differentiate 2 to the power x with respect to x?
The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.
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
What number to the power of itself equals 2 its something like 1.5596.... How is it calculated?
In order to answer this question, you need to know about natural logarithms and implicit differentiation as well as the Newton-Raphson (N-R) method for numerical approximation.Let y = x^xTake natural logarithms to give ln(y) = ln(x^x) = x*ln(x)Now differentiate: (1/y)*(dy/dx) = 1*ln(x) + x*1/x = ln(x) + 1Therefore, dy/dx = y*[ln(x) + 1] = x^x*[(ln(x) + 1]Now, solving for x^x = 2 is the same as solving for the root of x^x - 2 = 0. So let f(x) = x^x - 2 and use the N-R method to solve for its root.Start with an estimate x(0) = 2, say.Use the iteration x(n+1) = x(n) - f(x(n))/f'(x(n)) for n = 1, 2, 3, ...This gives x(4) = 1.559610563... and x(5) = 1.559610469 : sufficient accuracy for most people.
When you take the square root of a variable with an even number exponent how does the exponent value change?
When you take the square root of a variable raised to an exponent, you divide the exponent by two. For example the square root of x^4 is x^2, because x^2 x x^2 =x^4.
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
How do you differentiate 2 to the power x with respect to x?
The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.
Is logarithmic differentiation specially useful when dealing with functions with exponent that also depend on x?
Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x) Bring down the exponent. ln(y) = x ln(x^2 + 1) Differentiate both sides. dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Substitute in y = (x^2 + 1)^x. dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Solve for dy/dx. dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))
2 to the power of x equals 5 what is x?
To find the value of x when 2^x = 5, we can take the logarithm of both sides. Using the natural logarithm (ln) gives us: ln(2^x) = ln(5). Using the property of logarithms that allows us to bring the exponent down as a multiplier, we get: x*ln(2) = ln(5). Finally, dividing both sides by ln(2) gives us the value of x: x = ln(5)/ln(2), which is approximately 2.322.
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.
What is the logarythym equation 2 plus 3x equals 245?
Do you mean, 2 + 3^x = 245 ?? 2 + 3^x = 245 subtract 2 from each side 3^x = 243 use natural logs ( I always do, though you can use logs ) ln(3^x) = ln(243) you have the right in this operation to bring the exponent before the natural log x(ln3) = ln(243) divide x = ln(243)/ln(3) ( not ln(243/3) !!!! ) x = 5.493061443/1.098612289 x = 5 ----------check 2 + 3^5 = 245 2 + 243 = 245 245 = 245 ===========checks
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 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 anti-derivative of 5 to the x power?
ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)
What is the derivative of lnx2?
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/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 2 divided by x as a power of x?
x^(ln(2)/ln(x)-1)
Differentiate y equals a power x?
y=ax y'=ln(a)*ax
