0
e2x + ex = 0
e2x = -ex
e2x / ex = -1
ex = -1
x = ln(-1)
Actually, that's just a restatement of the original problem, but that's
as 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.
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Solve ex equals 2e1-2x?
meant to be e^x = 2e^1-2x
How does e raised to the power of 2x equal e raised to the power of x squared?
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
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 did ythey use to make coths?
coth(x) = cosh(x)/sinh(x) = (ex + e-x)/ (ex - e-x) or (e2x + 1)/ (e2x - 1)
What is e x squared?
If you mean ex squared, the answer is e2x
What is 6 ex takeaway3 plus 7 takeaway ex equals?
5x plus 4
Solve ex equals 2e1-2x?
meant to be e^x = 2e^1-2x
How does e raised to the power of 2x equal e raised to the power of x squared?
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
How do you prove ax5 plus bx4 plus cx3 plus dx2 plus ex plus f equals 0 is not solvable?
This is called the Abel-Ruffini theorem.
Solve e to the power of 2x plus 2e to power of x equals 8?
e2x + 2ex = 8. First, let's define z to be ex Substituting gives us the following quadratic equation:z2+2z=8 or z2+2z-8=0Solving for z gives us two possible zs: 2 and minus 4.As z is a power of e, only z=2 makes sense.Back solving we get:ex = 2and the answer is x=ln(2). If you substitute in the answer, you get e2ln2+2eln2=8, or 4+4=8. ln(2) ~= 0.69315.
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
If 2 plus 3 equals 10 and 7 plus 2 equals 63 and 6 plus 5 equals 66 and 8 plus 4 equals 96 what is 9 plus 7 equal too?
144 This works if you first add the two digits (ex. 2+3=5), then multiply the first digit by the answer you got (2+5=10).
What is the inverse of y equals to x plus 2 multiplied by e raised to the power of x?
Y = ex(x + 2) Y = ex/(X + 2) =========
E to the power of x SUBTRACT E to the power of -x divided by 2 equals 8 what is x?
(ex-e-x)/2=8 ex-e-x=16 ex = 16-e-x ln(ex) = ln(16-e-x) x = ln(16-e-x) x = ln((16ex-1)/ex) x = ln(16ex-1) - ln(ex) x = ln(16ex-1) - x 2x = ln(16ex-1) e2x = eln(16e^x - 1) e2x = 16ex-1 e2x-16ex +1 = 0 Consider ex as a whole to be a dummy variable "u". (ex=u) The above can be rewritten as: u2-16u+1=0 u2-16u=-1 Using completing the square, we can solve this by adding 64 to both sides of the equation (the square of one half of the single-variable coefficient -16): u2-16u+64=63 From this, we get: (u-8)2=63 u-8=(+/-)sqrt(63) u=sqrt(63)+8, u= 8-sqrt(63) Since earlier we used the substitution u=ex, we must now use the u-values to solve for x. sqrt(63)+8 = ex ln(sqrt(63)+8)= ln(ex) x = ln(sqrt(63)+8) ~ 2.769 8-sqrt(63) = ex ln(8-sqrt(63)) = ln(ex) x = ln(8-sqrt(63)) ~ -2.769 So, in the end, x~2.769 and x~-2.769. When backsubbed back into the original problem, this doesn't exactly solve the equation. Using a graphing calculator, the solution to this equation can be found to be approximately x=2.776 by graphing y=ex-e-x and y=16 and using the calculator to find the intersection of the two curves. This is pretty dang close to our calculated value, and rounding issues might account for this difference. The calculator, however, suggests that x~-2.769 is not a valid solution. This makes sense, and in fact it isn't a valid solution if you look at the graphs. This is an extraneous answer.
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.
