1. The problem statement, all variables and given/known data 3x^[tex]^{2}[/tex] = e^[tex]^{2x}[/tex] --- solve for x. 2. Relevant equations I burned it down to ln3 + 2lnx = 2x EDIT - you can also square root each side and get it down to x = ln([tex]\sqrt{3}[/tex]x) 3. The attempt at a solution ln(3x^2) = 2x ln3 + 2lnx = 2x (ln3)/2 = x - lnx I vaguely remember this problem earlier in the year. Yes, it can/should be done without a calculator, although I'm stuck at what I posted. There may be a special "trick" at the beginning that doesn't actually require this type of algebra. Alas, I cannot figure it out!
still working on it but here's what i got so far: [tex]\left( \sqrt{3}\,x-{e}^{x}\right) \,\left( \sqrt{3}\,x+{e}^{x}\right) =0[/tex] using cases(assuming if x- postive, or x is negative) I found that [tex]\left( \sqrt{3}\,x-{e}^{x}\right) =0[/tex] doesn't work and other one has a solution only when the x is close to 0, and is negative
Good so far. But I don't think there is any solution to this equation in terms of elementary functions. You just have to find the root numerically.
I mean that you have to use a technique like Newton's method to find an approximate root. There's no series of elementary functions like log's and exp's etc that you can punch in to get the answer.
You could use a physicists favorite method and take a second order taylor series. [tex]e^{2x} ~= 1 + 2x + 2x^2[/tex] Then solve the resulting quadratic. (Not sure how well that will work in this case)
Actually, I just received note that a calculator IS allowed on this problem. Sorry for the bother, but thanks for the help! EDIT - I would have probably used your methods if I was in Calculus :) (this is just the homework to get in)
I rather used linear approximation.. first, we know that value of x is negative and very close to 0. [e^x changes much rapidly than sqrt(3)*x function] and so using linear approximations.. i got -0.366025..which is close enough. i found equation of the tangent line at x=0 of e^x function and thus you would end up with a linear function with only one variable.
Newton Raphson zeroed in on -0.390646. Which when placed into each equation results in an equivalent number for both sides. It would be interesting to find an easier way to find a solution.
Using a calculator, I set each part of the equation to y1 and y2. So...in the y= section for graphing, y1 = 3x^2 y2 = e^2x Graph and then find the intersection of the two. I also got .-39 or something close to that.
Don't get too close with a graphics calculator. They're not really allowed in exams in a lot of places. Luckily over here they are. :) I'm quite interested to find out if there's a simple method you can use on paper and a pocket calculator to find a solution, other than bisection.
The solution to the equation [itex]0=\sqrt{3}x + e^x[/itex] can expressed in terms of the special function called the Lambert W function. Have a go at it; I'll supply help if you need it. Unfortunately, calculators don't have a W button like they do a sin button. This is the punchline to a joke. Question: What's the difference between an elementary function and a special function? Answer: an elementary is used so often that it's a button on a standard scientific calculator; a special function (for example, a Bessel function) often is used, but not so often that it's a button on a standard scientific calculator. Just as numerical algorithms are used to evaluate special functions, numerical algorithm are used to evaluate elementary function. We don't see this, because the numerical algorithms are hard-wired into our calculators. For example, to find [itex]sin(24.761)[/itex] (argument in radians), we just punch it into our cakculator. But what does the calculator do? It probably evalutes a suitable number of terms of [tex]sinx = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + ...[/tex] Now, for [itex]x=24.761[/itex], this series doesn't coverge very quickly, so the calculator probably "knows" (via symmetry properties) stuff like [itex]sin(24.761) = -sin(0.371)[/itex] (approximately), which, for a given accuracy, allows for evaluation of fewer terms in the series. Back to the original question. The solution can also be solved using the method of successive approximations, which is fairly easy to do by hand on a calculator. This is how I found x. This was particularly easy to automate on old programmable calculators. Drawbacks: 1) a little more cumbersome (maybe) to implement on today's programmable calculators; 2) convergence isn't guaranteed. Finally, the W function is one of Maple's built-in functions.