Germany | Can you solve this? | Math Olympiad

[@stpat7614]

Including the complex roots:
5^a * 5^a * 5^a = 60
5^(a + a + a) = (60^[1 / 3])^3
5^(3 * a) = ([2^2 * 3 * 5]^[1 / 3])^3
5^(a * 3) = ([2^2]^[1 / 3] * 3^[1 / 3] * 5^[1 / 3])^3
(5^a)^3 = (2^[2 * (1 / 3)] * 3^[1 / 3] * 5^[1 / 3])^3
(5^a)^3 = (2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3])^3
Let x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3)
x^3 = y^3
x^3 - y^3 = y^3 - y^3
x^3 - y^3 = 0
(x - y)(x^2 + xy + y^2) = 0
(x - y)(1 * x^2 + y * x + y^2) = 0
x - y = 0, or 1 * x^2 + y * x + y^2 = 0
Suppose x - y = 0
x - y = 0
x - y + y = 0 + y
x = y
Remember, x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3)
5^a = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3)
log(5^a) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3])
a * log(5) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3])
a * log(5) / log(5) = log(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3]) / log(5)
a * log_5(5) = log_5(2^[2 / 3] * 3^[1 / 3] * 5^[1 / 3])
a * 1 = log_5(2^[2 / 3]) + log_5(3^[1 / 3]) + log_5(5^[1 / 3])
a = (2 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5)
a = 2 * log_5(2) / 3 + 1 * log_5(3) / 3 + (1 / 3) * 1
a = 2 * log_5(2) / 3 + log_5(3) / 3 + (1 / 3)
Suppose 1 * x^2 + y * x + y^2 = 0
1 * x^2 + y * x + y^2 = 0
x = (-y +/- sqrt[y^2 - 4 * 1 * y^2]) / (2 * 1)
x = (-y +/- sqrt[1 * y^2 - 4 * y^2]) / (2)
x = (-y +/- sqrt[(1 - 4) * y^2]) / 2
x = (-y +/- sqrt[(-3) * y^2]) / 2
x = (-y +/- sqrt[(-1) * 3 * y^2]) / 2
x = (-y +/- sqrt[-1] * sqrt[3] * sqrt[y^2]) / 2
x = (-y +/- i * 3^[1 / 2] * y) / 2^(3 / 3)
x = 2^(-3 / 3) * y * (-1 +/- i * 3^[1 / 2] * 1)
Remember, x = 5^a, and y = 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3)
5^a = 2^(-3 / 3) * 2^(2 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2])
5^a = 2^([-3 / 3] + [2 / 3]) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 +/- i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2])
or
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2])
Suppose 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 + i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (i^2 + i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * i * (i + 1 * 3^[1 / 2])
ln(5^a) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)])
a * ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)])
a * ln(5) / ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i * [i + 3^(1 / 2)]) / ln(5)
a * log_5(5) = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(i) / ln(5) + ln(i + 3^[1 / 2]) / ln(5)
a * 1 = log_5(2^[-1 / 3]) + log_5(3^[1 / 3]) + log_5(5^[1 / 3]) + ln(e^[i * tau / 4]) / ln(5) + log_5(i + 3^[1 / 2])
a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5) + (i * tau / 4) * ln(e) / ln(5) + log_5(i + 3^[1 / 2])
a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * 1 + (i * tau / 4) * 1 / ln(5) + log_5(i + 3^[1 / 2])
a = -1 * log_5(2) / 3 + 1 * log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2])
a = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2])
Suppose 5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1 - i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * (-1) * (1 + i * 3^[1 / 2])
5^a = 2^(-1 / 3) * 3^(1 / 3) * 5^(1 / 3) * i^2 * (1 + i * 3^[1 / 2])
ln(5^a) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)])
a * ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)])
a * ln(5) / ln(5) = ln(2^[-1 / 3] * 3^[1 / 3] * 5^[1 / 3] * i^2 * [1 + i * 3^(1 / 2)]) / ln(5)
a * log_5(5) = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(i^2) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5)
a * 1 = ln(2^[-1 / 3]) / ln(5) + ln(3^[1 / 3]) / ln(5) + ln(5^[1 / 3]) / ln(5) + ln(e^[i * tau / 2]) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5)
a = (-1 / 3) * ln(2) / ln(5) + (1 / 3) * ln(3) / ln(5) + (1 / 3) * ln(5) / ln(5) + (i * tau / 2) * ln(e) / ln(5) + ln(1 + i * 3^[1 / 2]) / ln(5)
a = (-1 / 3) * log_5(2) + (1 / 3) * log_5(3) + (1 / 3) * log_5(5) + (i * tau / 2) * 1 / ln(5) + log_5(1 + i * 3^[1 / 2])
a = -1 * log_5(2) / 3 + 1 * log_5(3) / 3 + 1 * log_5(5) / 3 + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2])
a = -log_5(2) / 3 + log_5(3) / 3 + log_5(5) / 3 + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2])
a = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2])
a1 = 2 * log_5(2) / 3 + log_5(3) / 3 + (1 / 3)
a2 = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (4 * ln[5]) + log_5(i + 3^[1 / 2])
a3 = -log_5(2) / 3 + log_5(3) / 3 + (1 / 3) + i * tau / (2 * ln[5]) + log_5(1 + i * 3^[1 / 2])

[@zawatsky]

5^(a-1)=4=2²⇔lb(5^(a-1))=lb(2²)=2lb2=2=(a-1)lb5⇔a-1=2/lb5⇔a=1+2/lb5.

[@louisvanrijn3964]

Please comment: 3 . 5 ^a = 60 ; 5 ^a =20 = 4 .5 ; 5log5 ^a = 5log (4 . 5) ; a = 5log4 +1 ??? OK??

[@magalypolleunis7589]

Not sure but since log (a*b) = log a + log b if Im not mistaken it should instead be:
5log 5^a = 5log4 + 5log 5
=> 5a*log 5 = 5log4 + 5log5
=> a = [5log(4)/5log (5)] + 5log (5)/log(5)
=> a = log4/log5 + 1 = log5(4) + 1
Maybe simplier to use log 5 and not 5log5:
5^a =20 => log (5^a) = log(20) => a*log5 = log20
=> a = log(20)/log(5) => a = log5(20)
=> a = log5(5) + log5(4) = a = log5(4) + 1
a≈1.86

[@AlexanderSemashkevich]

x=log60/3log5=(log5+log4+log3)/3log5=⅓+(2log2+log3)/3log5
x≈⅓+(2×0.301+0.477)/(3×0.699)≈1778/2097≈0.848
Deviation is about 0.0001
5^(3×0.848)≈60.004