Math Olympiad: 2ˣ + 4ˣ + 8ˣ = 155; x = ? 2ˣ > 0; No complex and/or imaginary value root 2ˣ + 4ˣ + 8ˣ = 2ˣ + 2²ˣ + 2³ˣ = 2ˣ + (2ˣ)² + (2ˣ)³ = 155 Let: y = 2ˣ, 2ˣ + 4ˣ + 8ˣ = y + y² + y³ = 155, y³ + y² + y - 155 = 0; y > 0 (y³ - 125) + (y² - 25) + (y - 5) = (y³ - 5³) + (y² - 5²) + (y - 5) (y - 5)(y² + 5y + 25) + (y - 5)(y + 5) + (y - 5) = (y - 5)(y² + 6y + 31) = 0 y² + 6y + 31 = 0, (y - 3)² = - 22 = (i√22), y = 3 ± i√22 2ˣ = y = 3 ± i√22; Rejected, complex value root y - 5, y = 5 = 2ˣ, x = log₂5 = 2.322 The calculation was achieved on a smartphone with a standard calculator app Answer check: x = log₂5, 2ˣ = 5 2ˣ + 4ˣ + 8ˣ = 2ˣ[1 + 2ˣ + (2ˣ)²] = 5(1 + 5 + 25) = 155; Confirmed Final answer: x = log₂5 = 2.322

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I had difficulty following your logic after you already knew y=5. I also guessed 5 since it was a factor of 155, but then I did synthetic division to get: y^2+6y+31. This gave 2 complex answers which one was rejected. I need to understand your method better so that I don't have to resort to synthetic division all the time.

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by faktoring , let u=2^x , u^3+u^2+u-155=0 , 155=5*31 , -> u^2(u-5)+6u(u-5)+31(u-5)=0 , (u-5)(u^2+6u+31)=0 , u-5=0 , u=5 , 2^x=5 , x=ln5/ln2 , /// for complex , u^2+6u+31=0 , /// , test , 2^(ln5/ln2)+4^(ln5/ln2)+(ln5/ln2)=5+25+125 , 5+25+125 =155 , OK ,

I did it, faster, less steps actually

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