A Nice Algebra Problem | Math Olympiad | Can you solve it for x?

[@stpat7614]

With complex roots (I hope this is correct): 8^x - 2^x = 120 (2^3)^x - 2^x = 120 2^(3x) - 2^x = 120 2^(x*3) - 2^x = 120 (2^x)^3 - 2^x = 120 Let u = 2^x u^3 - u = 120 u^3 - u - 120 = 120 - 120 u^3 - u - 120 = 0 u^3 + (-25u + 24u) - (24 * 5) = 0 (u^3 - 25u) + (24u - [24 * 5]) = 0 u(u^2 - 25) + 24(u - 5) = 0 u(u^2 - 5^2) + 24(u - 5) = 0 u(u + 5)(u - 5) + 24(u - 5) = 0 (u - 5)(u * [u + 5] + 24) = 0 (u - 5)(u^2 + 5u + 24) = 0 Suppose u - 5 = 0 u - 5 = 0 u - 5 + 5 = 0 + 5 u = 5 Remember, u = 2^x 2^x = 5 log(2^x) = log(5) x * log(2) = log(5) x * log(2) / log(2) = log(5) / log(2) x = log_2(5) x1 = log_2(5) Suppose u^2 + 5u + 24 = 0 1 * u^2 + 5 * u + 24 = 0 Let a = 1, b = 5, c = 24 u = (-b +/- sqrt[b^2 - 4 * a * c]) / (2 * a) u = (-5 +/- sqrt[5^2 - 4 * 1 * 24]) / (2 * 1) u = (-5 +/- sqrt[25 - 96]) / (2) u = (-5 +/- sqrt[-71]) / 2 u = (-5 +/- sqrt[71 * (-1)]) / 2 u = (-5 +/- sqrt[71] * sqrt[-1]) / 2 u = (-5 +/- sqrt[71] * i) / 2 Remember, u = 2^x 2^x = (-5 +/- sqrt[71] * i) / 2 ln(2^x) = ln([-5 +/- sqrt(71) * i] / 2) x * ln(2) = ln([-5 +/- sqrt(71) * i] / 2) x * ln(2) / ln(2) = ln([-5 +/- sqrt(71) * i] / 2) / ln(2) x * 1 = ln([-5 +/- sqrt(71) * i] / 2) / ln(2) x = ln([-5 + sqrt(71) * i] / 2) / ln(2), or x = ln([-5 - sqrt(71) * i] / 2) / ln(2) x2 = ln([-5 + sqrt(71) * i] / 2) / ln(2) x2 = log_2([-5 + sqrt(71) * i] / 2) x2 = log_2(-5 + sqrt[71] * i) - log_2(2) x2 = log_2(-5 + sqrt[71] * e^[i * tau / 4]) - 1 x3 = ln([-5 - sqrt(71) * i] / 2) / ln(2) x3 = ln(-1 * [5 + sqrt(71) * i] / 2) / ln(2) x3 = ln(-1) / ln(2) + ln([5 + sqrt(71) * i] / 2) / ln(2) x3 = ln(e^[i * tau / 2]) / ln(2) + log_2([5 + sqrt(71) * i] / 2) x3 = (i * tau / 2) * ln(e) / ln(2) + log_2(5 + sqrt[71] * i) - log_2(2) x3 = (i * tau / 2) * 1 / ln(2) + log_2(5 + sqrt[71] * e^[i * tau / 4]) - 1 x3 = i * tau / (2 * ln[2]) + log_2(5 + sqrt[71] * e^[i * tau / 4]) - 1 {x1, x2, x3} = { log_2(5), log_2(-5 + sqrt[71] * e^[i * tau / 4]) - 1, i * tau / (2 * ln[2]) + log_2(5 + sqrt[71] * e^[i * tau / 4]) - 1 }

[@RichieSarabia-ms2er]

Outstanding Solution 👌👌👌

[@SALogics]

Very nice! ❤

[@RichieSarabia-ms2er]

Thank you so much for more uploads youtube videos like mathematics

[@SALogics]

More to come! ❤

[@RichieSarabia-ms2er]

Godbless You sir watching from Philippines 🙏🙏🙏

[@SALogics]

Thank you so much! ❤

[@daniellerosalie2155]

I like this problem.

[@ThienPhuc-FF]

Let 2^x=y Y³-y=120 Y³-y-120=0 Y³-y-125+5=0 (Y³-5³)-(y-5)=0 (Y-5)(y²+5y+25)-(y-5)=0 (Y-5)(y²+5y+24)=0 Case 1 Y-5=0 Y=5 2^x=5 Log2^x=log5 Xlog2=log5 X=log5 ² Case 2 Y²+5y+24=0 Delta=5²-4.1.24=25-96 =-71

[@walterwen2975]

Math Olympiad: 8ˣ - 2ˣ = 120; x =? 8ˣ - 2ˣ - 120 = 0, [(2ˣ)³ - 5³] - (2ˣ - 5) = (2ˣ - 5)[(2ˣ)² + 5(2ˣ) + 25 - 1] = 0 (2ˣ - 5)[(2ˣ)² + 5(2ˣ) + 24] = 0, 2ˣ > 0; (2ˣ)² + 5(2ˣ) + 24 > 0 2ˣ - 5 = 0, 2ˣ = 5, x = log₂5 = 2.322 The calculation was achieved on a smartphone with a standard calculator app Answer check: x = log₂5 = 2.322, 2ˣ = 5: 8ˣ - 2ˣ = (2ˣ)[(2ˣ)² - 1] = 5(25 - 1) = 120; Confirmed Final answer: x = log₂5 = 2.322

[@SALogics]

Very nice! ❤