It's been a while since I've been on sciforums, but recently, when I was tutoring, a student presented me this problem. It was a bonus problem that not even their teachers were able to solve, and it's got me stumped too! I figured it'd be a good one to share here... 2*x = 2^x, solve for x The answer is x=1,2, but I simply cannot get to those values algebraically. I've tried every form of logarithmic gymnastics I can think to do, and even plugged it into Wolfram Alpha out of desperation only to have it produce a numerically approximate solution. Anyone got any bright ideas? :shrug:

The equation reduces to : \(2^{x-1}=x\) This is a transcendental equation that is solved graphically. The LHS is an exponential, the RHS is the bisector of the first quadrant, so it can intersect the exponential in two points and two points only. These points are x=1 and x=2 , determined through direct observation. Thus your solution is complete. If you are really ambitious, you can show that there cannot be any solution for x>2 because \(f(x)=2^{x-1}-x\) is monotonically increasing. A similar mechanism is used to show that there cannot be any solution for 0<x<1.

I had that: 2 * x = 2^x 2^1 * x = 2^x x = 2^x / 2^1 x = 2^(x-1) There is no other way to solve it further than graphically? What about using log? I'm not big at math (understatement)..

It is a transcendental equation, such equations are solved through a combination of calculus and graphs as I have shown. There is a certain amount of calculus necessary. If you use the log you get an equivalent transcendental equation: \(x-1=log_2(x)\)

Thanks! Blows my mind that this can't be proven algebraically, but I'll definitely look into transcendental equations, I'm not exactly sure what that term even means! What a cruel thing to give a high-schooler for a bonus problem Please Register or Log in to view the hidden image!

Despite the impression given to high schoolers most maths problems cannot be solved algebraically. For instance, although there is a quadratic formula for the roots of \(ax^{2}+bx+c\) and likewise for cubic and quartic polynomials, there is no such equation for general quintics or higher. Not all functions can be integrated, either in the sense of having a closed form expression or in the sense of even having a well defined integral. An example of the former is \(f(x) = e^{-x^{2}}\), in that \(F(x) = \int^{x}_{-\infty}f(t)dt\) has no closed form expression in terms of nice functions (trig, polynomials, logs, exponentials etc). An example of the latter is a function which is 1 if x is rational and 0 if x is irrational. You can't even graph that one.

The same can be said about differential equations, only a very small subset has closed solutions. Closed solutions are more the exception than the rule.