Can you crack this beautiful equation? - Harvard exam question

[@davidseed2939]

(4^a)^3=4*11
4^(3a-1) =11
3a-1=log_4(11)
a= (1+log_4(11))/3

[@doublestone1]

15 Seconds - not 8:40 ...

[@jjjilani9634]

Use log from the beginning. Log4^3a=log44. a=.91

[@abeonthehill166]

I think you complicated this equation man, thanks for sharing .

[@stpat7614]

Aren't there also two complex roots?
4^a*4^a*4^a = 44
(4^a)^3 = 44
(4^a)^3 - 44 = 44 - 44
(4^a)^3 - 44 = 0
(4^a)^3 - (44^[1/3])^3 = 0
Let u = 4^a
u^3 - (44^[1/3])^3 = 0
(u - 44^[1/3])(u^2 + u*44^[1/3] + [44^(1/3)]^2) = 0
(u - 44^[1/3])(u^2 + u*44^[1/3] + 44^[(1/3)*2]) = 0
(u - 44^[1/3])(u^2 + 44^[1/3]*u + 44^[2/3]) = 0
u - 44^(1/3) = 0, or 1*u^2 + 44^(1/3)*u + 44^(2/3) = 0
1*u^2 + 44^(1/3)*u + 44^(2/3) = 0
Let a = 1, b = 44^(1/3), c = 44^(2/3)
u = (-b +/- sqrt[b^2 - 4*a*c]) / (2*a)
u = (-44^[1/3] +/- sqrt[(44^[1/3])^2 - 4*1*44^(2/3)]) / (2*1)
u = (-44^[1/3] +/- sqrt[44^(1/3*2) - 4*44^(2/3)]) / 2
u = (-44^[1/3] +/- sqrt[44^(2/3) - 4*44^(2/3)]) / 2
u = (-44^[1/3] +/- sqrt[44^(2/3)*(1 - 4)]) / 2
u = (-44^[1/3] +/- sqrt[44^(2/3)*(-3)]) / 2
u = (-44^[1/3] +/- sqrt[44^(2/3)*3*(-1)]) / 2
u = (-44^[1/3] +/- sqrt[44^(2/3)]*sqrt[3]*sqrt[-1]) / 2
u = (-44^[1/3] +/- [44^(2/3)]^[1/2]*3^[1/2]*[-1]^[1/2]) / 2
u = (-44^[1/3] +/- 44^[(2/3)*(1/2)]*3^[1/2]*i) / 2
u = (-44^[1/3] +/- 44^[1/3]*3^[1/2]*i) / 2
u = (-1 +/- 1*3^[1/2]*i)*44^(1/3) / 2
u = (-1 +/- 3^[1/2]*i)*(2^2*11)^(1/3) / 2
u = (-1 +/- 3^[1/2]*i)*2^(2/3)*11^(1/3) / 2^(3/3)
u = (-1 +/- 3^[1/2]*i)*2^([2/3] - [3/3])*11^(1/3)
u = (-1 +/- 3^[1/2]*i)*2^(-1/3)*11^(1/3)
Remember u = 4^a
4^a = (-1 +/- 3^[1/2]*i)*2^(-1/3)*11^(1/3)
log(4^a) = log([-1 +/- 3^(1/2)]*i]*2^[-1/3]*11^[1/3])
a*log(4) = log([-1 +/- 3^(1/2)]*i]*2^[-1/3]*11^[1/3])
a*log(4)/log(4) = log([-1 +/- 3^(1/2)]*i]*2^[-1/3]*11^[1/3]) / log(4)
a*1 = log([-1 +/- 3^(1/2)]*i]*2^[-1/3]*11^[1/3]) / log(2^2)
a = log([-1 +/- 3^(1/2)]*i]*2^[-1/3]*11^[1/3]) / (2*log[2])
a = (log[-1 +/- 3^(1/2)]*i] + log[2^(-1/3)] + log[11^(1/3)]) / (2*log[2])
a = (log[-1 +/- 3^(1/2)]*i] - 1/3 * log[2] + 1/3*log[11]) / (2*log[2])
a = (3*log[-1 +/- 3^(1/2)]*i]/3 - 1/3 * log[2] + 1/3*log[11]) / (2*log[2])
a = (3*log[-1 +/- 3^(1/2)]*i] - log[2] + log[11]) / (3*2*log[2])
a = (3*log[-1 +/- 3^(1/2)]*i] + log[11] - log[2]) / (6*log[2])
a = (3*log_2[-1 +/- 3^(1/2)]*i] + log_2[11] - log_2[2]) / 6
a = (3*log_2[-1 +/- 3^(1/2)]*i] + log_2[11] - 1]) / 6
a = (3*log_2[-1 + 3^(1/2)]*i] + log_2[11] - 1]) / 6, or a = (3*log_2[-1 - 3^(1/2)]*i] + log_2[11] - 1]) / 6
a = (3*log_2[i^2 + 3^(1/2)]*i] + log_2[11] - 1]) / 6, or a = (3*log_2[i^2 + i^2*3^(1/2)]*i] + log_2[11] - 1]) / 6
a = (3*log_2[i(i + 3^[1/2])] + log_2[11] - 1]) / 6, or a = (3*log_2[i^2*(1 + 3^[1/2]*i)] + log_2[11] - 1]) / 6
a = (3*log_2[i] + 3*log_2[i + 3^(1/2)] + log_2[11] - 1]) / 6, or a = (3*log_2[i^2] + 3*log_2[1 + 3^(1/2)*i] + log_2[11] - 1]) / 6
a = (3*log_2[i] + 3*log_2[i + 3^(1/2)] + log_2[11] - 1]) / 6, or a = (6*log_2[i] + 3*log_2[1 + 3^(1/2)*i] + log_2[11] - 1]) / 6
a1 = (2 + log_2[11])/6
a2 = (3*log_2[i] + 3*log_2[i + 3^(1/2)] + log_2[11] - 1]) / 6
a3 = (6*log_2[i] + 3*log_2[1 + 3^(1/2)*i] + log_2[11] - 1]) / 6

[@doublestone1]

I hate these click-bait titles: "99% of students failed". Sorry this is something what a bull sees, when he looks behind. There is absolutely nothing difficult with that equation.

[@RealQinnMalloryu4]

{11+11+11}=33 11^11^11 5^6^5^6^5^6 2^3^3^3^2^3^3^3^2^3^3^3 1^1^1^1^1^1^1^1^2^1^1^3 23(a ➖ 3a+2).