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How do you solve for the power of x? For example, 2^x = 423 2 Oct 2014 · 2^x=423 Take the natural log of both sides ln (2^x)= ln (423) Use one of properties of logs to move the exponent down as a factor x*ln (2)=ln(423) Use Algebra to solve for x by …
What is the integral of #e^(x^3)#? - Socratic 5 Jan 2015 · This means that its primitive functions are F:\mathbb{R} to \mathbb{R} such that F(y) = c + int_0^y e^{x^3}dx=c- 1/3 int_0^{-y^3} e^{-t} t^{-2/3} dt and this is well defined because the …
What is 2 to the power -3? - Socratic 6 Dec 2017 · The answer is 1/8. Since your exponent is negative, you will have to make a fraction. To make the exponent even, you will make 1 the numerator and 2 to the power of 3 the …
How do you find the Maclaurin Series for #Sin(x^2)#? - Socratic 12 Nov 2017 · x^2 - x^6/(3!) + x^10/(5!) - .... sum_(n=0 )^oo x^(4n+2)/((2n+1)!) * (-1)^n First we must find the series for sin(x) let f(x) = sin(x) f(0) = sin(0) = 0 f'(0) = cos(0 ...
How do you simplify 4 to the power of 2/3? - Socratic 14 Apr 2015 · Remember that a^(m/n)=rootn(a^m) and that 4^2=16=2xx8=2xx2^3 ...the rest is a piece of cake!
How do you write x^(2/3) in radical form? - Socratic 9 Aug 2017 · First, we can rewrite the term as: #x^(2 xx 1/3)# Next, we can use this rule of exponents to rewrite the term again:
What is the remainder of 3^29 divided by 4? - Socratic 25 Feb 2018 · #3^29/4# when 3^0 =1 is divided by 4, the remainder is 1 when 3^1 =3 is divided by 4, the remainder is 3 when 3^2 =9 is divided by 4, the remainder is 1 when 3^3 =27 is divided …
How do you simplify #25^(3/2)#? - Socratic 3 Feb 2015 · The number reads as 25 to the power of (3/2), which simplifies to sqrt(15625). When an exponent is a fraction, the denominator of the fraction becomes the root. We can take the …
How do you simplify #9^(3/2)#? - Socratic 25 May 2016 · 27 Breaking 9^(3/2) down into its component parts 3/2 is the same as 3xx1/2 which is the same as 1/2xx3 so 9^(3/2) is the same as 9^(1/2xx3) Write this as: (9^(1/2))^3 Consider …
Power Rule - Calculus - Socratic #y^' = nx^(n-1)# Below are the proofs for every numbers, but only the proof for all integers use the basic skillset of the definition of derivatives.