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How do you solve for x where e2x plus ex equals 0?
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas far as you can go with it, because there's no such thing as the log(or the ln) of a negative number. Your equation has no solution.
Solve e to the power x equals x to the power 3 algebraically?
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.
Why was the calculus student confused about y equals ex and the derivative of y equals ex?
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Y' equals x plus y Explicit solution?
using Laplace transform, we have: sY(s) = Y(s) + 1/(s2) ---> (s-1)Y(s) = 1/(s2), and Y(s) = 1/[(s2)(s-1)]From the Laplace table, this is ex - x -1, which satisfies the original differential eq.derivative of [ex - x -1] = ex -1; so, ex - 1 = ex - x - 1 + xto account for initial conditions, we need to multiply the ex term by a constant CSo y = C*ex - x - 1, and y' = C*ex - 1, with the constant C, to be determined from the initial conditions.
What is the integral of e raised to x cubed?
(ex)3=e3x, so int[(ex)3dx]=int[e3xdx]=e3x/3 the integral ex^3 involves a complex function useful only to integrations such as this known as the exponential integral, or En(x). The integral is:-(1/3)x*E2/3(-x3). To solve this integral, and for more information on the exponential integral, go to http://integrals.wolfram.com/index.jsp?expr=e^(x^3)&random=false
Solve e raised to the power of x equals 6?
If you want to solve ex =6, you need to take the natural log of both sides.ln ex =ln6Now we have x=ln6 which can be left that way or approximated with a calculator.ln means natural log, in case there was any question about that.The answer is about 1.79176.
How do you solve integer math problems?
If you're adding: Ex. -3 + 15 (-3, -2, -1, 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) = 12If you're subtraction: Ex. -17 - 20 (add the opposite, so problem really equals 17+20)= 37If you're multiplying: Ex. 5 x -10 (one negative equals negative answer. both positive or negative equals positive answer.)= -50If you're dividing: Ex. -100/10 (same rules as multiplication)= -10
How do you solve for x where e2x plus ex equals 0?
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas far as you can go with it, because there's no such thing as the log(or the ln) of a negative number. Your equation has no solution.
If ex equals 0 x equals?
x=0
Solve e to the power x equals x to the power 3 algebraically?
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.
Solve e to the negative x?
I am not sure what you want to "solve" here. e-x is the same as 1 / ex.
What is 6 ex takeaway3 plus 7 takeaway ex equals?
5x plus 4
Solve for x 9e3x equals 27ex-1 x-1 and 3x are exponents?
9 e3x = 27 ex-1e3x / ex-1 = 27/9 = 3e(3x - x + 1) = 3e2x+1 = 32x + 1 = ln(3) = 1.098612x = 0.09861x = 0.09861 / 2 = 0.04931(rounded)
How do you answer pie?
You don't answer it you use it to solve problems with EX: 3+3.14=6.14
How do you solve an equation with a variable on each side?
It's not always to 'solve' as it is to 'simplify' EX: 15a + 3b = 45c simplifies to 5a + b = 15c
How to solve e to the power of 3x plus 5e to the power of x equals 7?
e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4
Why was the calculus student confused about y equals ex and the derivative of y equals ex?
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
