This was a fun one to solve. I used u^6 = x, so u^3 = sqrt(x) and u^2 = cbrt(x). Wound up with a quartic 3u^4 - 2u^2+3u^2=0. Factored out u^2 (x = 0 is an obvious solution) and we're left with a quadratic where u = 1/3 +/- (2sqrt(2)/3)i. Raising this to the 6th power was the worst part, but I got x = 329/729 +/- (460sqrt(2)/729)i. SO three distinct solutions (because 0 came up twice). Crossing my fingers on this one. EDIT: yeah, I double-checked my complex solution and it's messed up. Not sure why. Oh well my brain is tired.

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problem 3ˣ - (2√2)ˣ = 1 let u = 3ˣ and v = (2√2)ˣ We know u-v = 1. That is the main equation with which we are dealing. Therefore u-v-1 = 0 Set x = x by setting ln u/ ln 3 = ln v / ln (2√2) ln u/ ln 3 = ln v / ln (2√2) Replace u = v+1 ln (2√2)/ ln 3 = ln v / ln (v+1) ln v / ln (v+1) = ln (2√2)/ ln 3 The only solution is v=8 so ln v / ln (v+1) = ln 8 / ln 9 = 3 ln 2 / 2 ln 3 = ln (2³ᐟ²)/ln 3 u= 9=3ˣ v = 8= (2√2)ˣ answer x = 2

Something went wrong with your extended analysis. I thought x = 1 is also a solution and your Wolfram Alpha graphs must be based on the remainder polynomial of x=0 as the only other real solution!🤣 At 3:24 you were first talking about a try solution of y^2 = 1 and y = +/- 1 from that gave the y = 1 and the extraneous other solution y = -1 other solution. I think you should have discussed factoring your first x = y^6 - 2y^3 + 1 into (y - 1) ( quadratic of a square remainder) to know y = 1 solves before the 3:24 mark)!!