An Exponential Equation In Two Variables | Real Solutions?

[@spelunkerd]

What an interesting approach. it made me pause when you made the creative assumption that Y and X are linearly related. That assumption gives a clear solution to the first problem, but does that method exclude other reasonable solutions, too? After all, what if X and Y are instead exponentially related, or related by a quadratic? Is it fair to assume linearity?

[@trojanleo123]

One equation two unknowns without any other information by the laws of Algebra cannot be solved. The assumption y = kx assumes the two variables are linearly related which may not be true.

[@MohammedRasheed-d8m]

graphing the equation in Desmos does not give the same graph as the one you have included in your video. why so?

[@alexandermorozov2248]

Yes, I also got completely different graphs if I build the functions x(k) and y(k).

[@alexandermorozov2248]

P. S. This is a graph defined parametrically - in the form of x(t) and y(t). The graph has two branches - the lower one (t from 0 to 2) and the upper one (t from 2 to infinity).

[@Don-Ensley]

problem xʸ = y²ˣ Set y = k x, k ∈ ℝ xᵏˣ = ( k x )²ˣ Take natural logs on each side. k x ln x = 2x ln(k x ) k x ln x = 2 x ln k + 2 x ln x ( k - 2 ) x ln x - 2 x ln k = 0 x [ ( k - 2 ) ln x - 2 ln k ] = 0 Use the 0 product property. x = 0, y = 0 is one solution. ( k - 2 ) ln x - 2 ln k = 0 x = k ²ᐟ⁽ᵏ⁻²⁾ y = k ᵏᐟ⁽ᵏ⁻²⁾ This is an interesting parametric curve with parameter k. It has 2 separate branches and a bump where the slope is 0 on the first branch at the local maximum and a valley where the slope is 0 at the local minimum of the second branch. The slope dy/dx may be calculated. dy/dx = (dy/dk ) / (dx/dk) y = k ᵏᐟ⁽ᵏ⁻²⁾ ln y = ln k k /(k-2) 1/y dy/dk = 1/(k-2) + ln k (-2)/[(k-2)²] dy/dk = k ᵏᐟ⁽ᵏ⁻²⁾ ( k -2 -2 ln k ) / [(k-2)² ] x = k ²ᐟ⁽ᵏ⁻²⁾ ln x = 2 ln k / (k-2) 1/x dx/dk = 2 [ (k-2) / k - ln k ] / [(k-2)² ] dx/dk = 2k ²ᐟ⁽ᵏ⁻²⁾ [ (k-2) / k - ln k ] / [(k-2)² ] dy/dx = (k/2) ( k -2 -2 ln k ) /[ (k-2) / k - ln k ] Setting this to 0, dy/dx = 0 at k = 0 and when k -2 -2 ln k = 0 k = 2(1 + ln k) Let u = 1 + ln k k = eᵘ⁻¹ eᵘ⁻¹ = 2 u eᵘ = 2e u 1/(2e) = u e⁻ᵘ -u e⁻ᵘ = -1/(2e) Use Lambert's product log to find where the slope is zero. - u = W[-1/(2e)] u = - W[-1/(2e)] = 1 + ln k ln k = -{ W[-1/(2e)] + 1 } k = e^(-( W₀(-1/(2e)) + 1 )) This is at a local maximum on branch 1. The corresponding x and y are: (x,y) = ( e, e^(- W₀(-1/(2e))) ) (x,y)≈ (2.71828, 1.26107) There are 2 values of k where the slope is zero because of the discontinuity at k = 2, the other branch also has a zero slope at k = e^(-( W₋₁(-1/(2e)) + 1 )) This is a local minimum on branch 2. The corresponding x and y are: ( e, e^(-W₋₁(-1/(2e)) ) ) (x,y) ≈ (2.71828, 14.561) Desmos graph: www.desmos.com/calculator/uuykdmhikf Wolfram Alpha command to plot is www.wolframalpha.com/input?i=parametric+plot+%28t%5E%282%2F%28t-2%29%29%2C++t%5E%28t%2F%28t-2%29%29%29 answer (x, y) = ( k ²ᐟ⁽ᵏ⁻²⁾, k ᵏᐟ⁽ᵏ⁻²⁾ ), k ∈ ℝ