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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
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 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.
What is 6x6x5x6x3x2x3 in exponent form?
6x6x5x6x3x2x3 in exponent form is 2 x 32 x 5 x 63
