What is 6e e as in exponential function?
e^x then e is Euler's number so 6e would be 6 * 2.718281828459... however on a calculator displayed 6eNN is a short hand for exponential notation 6 * 10^(NN)
Exponential function equation?
(a) V(t) = 2.033Volts, simply by plugging the numbers into the equation. (b) ln(1) = ln(5) - 0.3t t = 5.365seconds (c) dV/dt = Rate of Voltage fall = -1.5.e^(-0.3t). V(1)/V(0) = 3.70409/5 = 0.74081 Therefore, after 1st second, the voltage has decreased by 25.92% V(2)/V(1) = 2.744058/3.70409 = 0.74081 Therefore, during the 2nd second, the voltage has decreased by 25.92% also.
Exponential Functions?
Hello, It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks". Anyway... = = = = = = = = = = = = = = = = = = When x=ln(3) and y=ln(2): 6.e^(2x) + k.e^y + e^(2y) = c 6.e^[2.ln(3)] + k.e^[ln(2)] + e^[2.ln(2)] = c 6×3² + k×2 + 2² = c 54 + 2k + 4 = c 58 + 2k = c → QED Differentiating the equation with respect to x: 6.e^(2x) + k.e^y + e^(2y) = c 12.e^(2x) + k.y'.e^y + 2y'.e^(2y) = 0 When x=ln(3) and y=ln(2), then y'=-6: 12.e^(2x) + k.y'.e^y + 2y'.e^(2y) = 0 12.e^[2.ln(3)] + k×(-6)×e^[ln(2)] + 2×(-6)×e^[2.ln(2)] = 0 12×3² – 6k×2 – 12×2² = 0 108 – 12k – 48 = 0 9 – k – 4 = 0 → k = 5 Then: → c = 58 + 2k = 58 + 10 = 68 Regards, Dragon.Jade :-)
Exponential/Logarithmic Function?
Hi, I would start by b. b/ The population after 0 year was 150*10^6*e^0 = 150*10^6. Let's call it P(0) a/ The population after t years will be P(0)*(1.0296)^t And the problem says that this is equal to P(0)*e^kt Then, 1.0296^t = e^kt Then, 1.0296 = e^k k = ln 1.0296 = 0.0297 For c you can use either equation P(2000) = P(0)*e^(0.0297*50) or P(2000) = P(0)*(1.0296)^50 (1.0296)^50 = 4.29958 The population in 2000 will be 150*10^6*4.29958 = 644.9*10^6 d. P(x) = 2*P(0) ==> e^(0.0297*t) = 2 ==> ln 2 = 0.0297t 0,693 = 0,0297*t t = 0.693 / 0.0297 = 23.33 years Hope it helps Bye !!!
Help find dy/dx with exponential?
I'm going to assume you mean: e^(2y) = 28x² + 7y² Using implicit differentiation we have: 2e^(2y) dy/dx = 56x + 14y dy/dx e^(2y) dy/dx = 28x + 7y dy/dx (e^(2y) - 7y) dy/dx = 28x dy/dx = 28x/(e^(2y) - 7y) Hope this helps! ========== EDIT 1: I've corrected my answer above. However, right now I'm having a brain freeze and don't see how to convert back lol We could have done this problem differently and arrived at one of your options, but we'd have to use logarithmic differentiation. Take the natural logarithm of both sides: ln(e^(2y)) = ln(28x² + 7y²) 2yln(e) = ln(28x² + 7y²) 2y = ln(28x² + 7y²) Now implicitly differentiate: 2 dy/dx = (56x + 14y dy/dx)/(28x² + 7y²) 56x² dy/dx + 14y² dy/dx = 56x + 14y dy/dx 4x² dy/dx + y² dy/dx = 4x + y dy/dx 4x² dy/dx + y² dy/dx - y dy/dx = 4x (4x² + y² - y) dy/dx = 4x dy/dx = 4x/(4x² + y² - y) Hope this helps! ========== EDIT 2: There it is! And it was so simple! I can't believe I missed it! lol Notice that we have that e^(2y) = 28x² + 7y² Therefore, in my original answer I have: dy/dx = 28x/(e^(2y) - 7y) Substituting: dy/dx = 28x/([28x² + 7y²] - 7y) Now simplify and you'll get the same answer! My mind is at rest now lol :o) Hope this helps!
Exponential growth problem?
In 1999, the world's population reached 6 billion and was increasing at a rate of 1.3% per year. Assume that this growth rate remains constant. (In fact, the growth rate has decreased since 1987.) Write a formula for the world population (in bill.ions) as a function of the number of years, t, since 1999. Also, estimate the population for 2015
Solving exponential equations by taking the natural log of both sides?? Please help?
Part a) M = 6 e^0 M = 6 Part b) 90 = 6 e^(1.18 t ) 15 = e^(1.18 t ) ln 15 = 1.18 t ln e ln 15 = 1.18 t t = ln 15 / 1.18 t = 2.29 months
Calculus question spring function?
The derivative of a position function = The velocity function. The derivative of a velocity function = The acceleration function. So, to get your acceleration function, you need to take the second derivative of that function. This is a messy problem, since it involves the product rule: s(t) = 6e^(-2.2t) * sin(2pi * t) s'(t) = v(t) = (6e^(-2.2t) * (-2.2))(sin(2pi * t)) + (6e^(-2.2t))((cos(2pi * t) * (2pi)) v(t) = (-13.2sin(2pi * t) * e^(-2.2t)) + (12pi * cos(2pi * t) * e^(-2.2t)) Keep in mind that the derivative of e^(ax) = e^(ax) * a. Take the second derivative of the position function/the first derivative of the velocity function for your acceleration function. Apply the chain rule, the "e" rule, and the product rule for this one.
Endpoints of f (t) = ln ( 1 - 6e^-t ) ?
For f(t) = ln(1 - 6e^(-t)) to be defined we need the part inside the ln() to be strictly positive, i.e. f(t) is defined when 1 - 6e^(-t) > 0 <=> 6e^(-t) < 1 <=> e^(-t) < 1/6 <=> -t < ln (1/6) (since e^x is a strictly increasing function of x) <=> t > -ln (1/6) = - (- ln 6) = ln 6. So f(t) is defined on the open interval (ln 6, ∞).
Why is e more important than any other constant (e.g. 2)?
I am not sure it makes sense to compare the importance of constant in mathematics. Without the constant number 2, there would be no number e. But e is a strange looking irrational (and transcendent) number, with a definition which does not seem obvious without some study lim(n->infinity) of (1+1/n)^n. It differs from PI on that, as PI is simply the length of a circle divided by its diameter. Yet, like PI, we can meet e very often, and I guess you are asking why. The main reason why is related to the theory of derivatives. Take any logarithmic function, with the logarithmic base being any (positive) number a. The derivative of the function log_a(x), computed from the analytical definition of the derivative will give the function (1/x) multiplied by log_a(e). Similarly the derivative of any exponential function a^x will give a^x divided by the log_a(e). So the number e appears in each case. This implies also that the eulerian exponential e^x is a fixed point for the derivative (the derivative of e^x is itself), which makes it into a powerful tool in the theory of differential equations. The only other function equal to its own derivative is the constant function 0. That makes the number e very important in physics. Another reason of the importance of e si that the development in Taylor series of e^ix gives cos(x) + isin(x), so that any complex number can take the form re^it (I denotes the square root of (-1)). That is very handy to multiply complex numbers, and it simplifies trigonometry a lot, and thus it simplifies also a lot any study of vibration, waves, circular motions, etc.I could add the amazing fact that the harmonic series 1+(1/2)+(1/3)+ …, which diverges, behave like the log_e(x). In the limit, the series and the log_e get equal, modulo another constant, called gamma, due to Euler (and Mascheroni), which is about equal to 0,5. The constant gamma has been calculated and it looks like an irrational number, but that is a problem which is still open.