meant to be e^x = 2e^1-2x
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)
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
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.
coth(x) = cosh(x)/sinh(x) = (ex + e-x)/ (ex - e-x) or (e2x + 1)/ (e2x - 1)
If you mean ex squared, the answer is e2x
5x plus 4
meant to be e^x = 2e^1-2x
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)
This is called the Abel-Ruffini theorem.
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.
e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4
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).
Y = ex(x + 2) Y = ex/(X + 2) =========
(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.
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.