Germany | Can you solve this? | Math Olympiad

[@evrankorkut8915]

Thank you very much

[@stpat7614]

With complex roots (someone verify):
4^a * 4^a = 80
(4^a)^2 = 80
(4^a)^2 = (sqrt[80])^2
(4^a)^2 = (sqrt[2^4 * 5])^2
(4^a)^2 = (sqrt[2^4] * sqrt[5])^2
(4^a)^2 = ([2^4]^[1/2] * 5^[1/2])^2
(4^a)^2 = (2^[4 * (1/2)] * 5^[1/2])^2
(4^a)^2 = (2^2 * 5^[1/2])^2
Let x = 4^a, and y = 2^2 * 5^(1/2)
(4^a)^2 = (2^2 * 5^[1/2])^2
=> x^2 = y^2
=> x^2 - y^2 = y^2 - y^2
=> x^2 - y^2 = 0
=> (x - y)(x + y) = 0
=> (4^a - 2^2 * 5^[1/2])(4^a + 2^2 * 5^[1/2]) = 0
Remember, (4^a)^2 = (2^2 * 5^[1/2])^2
(4^a)^2 = (2^2 * 5^[1/2])^2
([2^2]^a)^2 = ([sqrt(2^2 * 5^[1/2])]^2)^2
(2^[2 * a])^2 = ([sqrt(2^2) * sqrt(5^[1/2])]^2)^2
(2^[a * 2])^2 = ([(2^2)^(1/2) * (5^[1/2])^(1/2)]^2)^2
([2^a]^2)^2 = ([2^(2 * [1/2]) * 5^([1/2] * [1/2])]^2)^2
([2^a]^2)^2 = ([2^(2/2) * 5^(1/4)]^2)^2
([2^a]^2)^2 = ([2 * 5^(1/4)]^2)^2
Let x = 2^a, and b = 2 * 5^(1/4)
(4^a - 2^2 * 5^[1/2])(4^a + 2^2 * 5^[1/2]) = 0
=> (x^2 - y^2)(x^2 + y^2) = 0
=> (x - y)(x + y)(x - y * i)(x + y * i) = 0
=> (2^a - 2 * 5^[1/4])(2^a + 2 * 5^[1/4])(2^a - 2 * 5^[1/4] * i)(2^a + 2 * 5^[1/4] * i) = 0
Let n be an integer
Suppose 2^a - 2 * 5^(1/4) = 0
2^a - 2 * 5^(1/4) + 2 * 5^(1/4) = 0 + 2 * 5^(1/4)
2^a = 2 * 5^(1/4)
log(2^a) = log(2 * 5^[1/4])
a * log(2) = log(2 * 5^[1/4])
a * log(2) / log(2) = log(2 * 5^[1/4]) / log(2)
a * 1 = log(2 * 5^[1/4]) / log(2)
a = log_2(2 * 5^[1/4])
a = log_2(2) + log_2(5^[1/4])
a = log_2(2) + log_2(5^[1/4])
a = 1 + (1/4) * log_2(5)
a = 1 + log_2(5) / 4
a1 = 1 + log_2(5) / 4
Suppose 2^a + 2 * 5^(1/4) = 0
2^a + 2 * 5^(1/4) = 0
2^a + 2 * 5^(1/4) - 2 * 5^(1/4) = 0 - 2 * 5^(1/4)
2^a = -2 * 5^(1/4)
ln(2^a) = ln(-2 * 5^[1/4])
a * ln(2) = ln(-2 * 5^[1/4])
a * ln(2) / ln(2) = ln(-2 * 5^[1/4]) / ln(2)
a * 1 = ln(-2 * 5^[1/4]) / ln(2)
a = ln(-1 * 2 * 5^[1/4]) / ln(2)
a = ln(-1) / ln(2) + ln(2 * 5^[1/4]) / ln(2)
a = ln(e^[i * n * tau / 2]) / ln(2) + log_2(2 * 5^[1/4])
a = (i * n * tau / 2) * ln(e) / ln(2) + log_2(2 * 5^[1/4])
a = (2 * i * n * tau / 2^2) * ln(e) / ln(2) + log_2(2) + log_2(5^[1/4])
a = (i * 2 * n * tau / 4) * 1 / ln(2) + 1 + (1/4) * log_2(5)
a = (i * 2 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a2 = (i * 2 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
Suppose 2^a - 2 * 5^(1/4) * i = 0
2^a - 2 * 5^(1/4) * i = 0
2^a - 2 * 5^(1/4) * i + 2 * 5^(1/4) * i = 0 + 2 * 5^(1/4) * i
2^a = 2 * 5^(1/4) * i
ln(2^a) = ln(2 * 5^[1/4] * i)
ln(2^a) = ln(i * 2 * 5^[1/4])
a * ln(2) = ln(i * 2 * 5^[1/4]) / ln(2)
a * ln(2) / ln(2) = ln(i * 2 * 5^[1/4]) / ln(2)
a * 1 = ln(i * 2 * 5^[1/4]) / ln(2)
a = ln(i) / ln(2) + ln(2 * 5^[1/4]) / ln(2)
a = ln(e^[i * n * tau / 4]) / ln(2) + log_2(2 * 5^[1/4])
a = (i * n * tau / 4) * ln(e) / ln(2) + log_2(2) + log_2(5^[1/4])
a = (i * n * tau / 4) * 1 / ln(2) + 1 + (1/4) * log_2(5)
a = (i * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a3 = (i * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
Suppose 2^a + 2 * 5^(1/4) * i = 0
2^a + 2 * 5^(1/4) * i = 0
2^a + 2 * 5^(1/4) * i - 2 * 5^(1/4) * i = 0 - 2 * 5^(1/4) * i
2^a = -2 * 5^(1/4) * i
ln(2^a) = ln(-2 * 5^[1/4] * i)
a * ln(2) = ln(-1 * i * 2 * 5^[1/4])
a * ln(2) / ln(2) = ln(-1 * i * 2 * 5^[1/4]) / ln(2)
a * 1 = ln(-1 * i * 2 * 5^[1/4]) / ln(2)
a = ln(-1 * i) / ln(2) + ln(2 * 5^[1/4]) / ln(2)
a = ln(-1 * i) / ln(2) + log_2(2 * 5^[1/4])
a = ln(e^[i * n * tau / 2] * e^[i * n * tau / 4]) / ln(2) + log_2(2) + log_2(5^[1/4])
a = ln(e^[(i * n * tau / 2) + (i * n * tau / 4)]) / ln(2) + 1 + (1/4) * log_2(5)
a = ln(e^[(i * 2 * n * tau / 2^2) + (i * n * tau / 4)]) / ln(2) + 1 + (1/4) * log_2(5)
a = ln(e^[i * 3 * n * tau / 4]) / ln(2) + 1 + (1/4) * log_2(5)
a = (i * 3 * n * tau / 4) * ln(e) / ln(2) + 1 + log_2(5) / 4
a = (i * 3 * n * tau / 4) * 1 / ln(2) + 1 + log_2(5) / 4
a = (i * 3 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a4 = (i * 3 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a1 = 1 + log_2(5) / 4
a2 = (i * 2 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a3 = (i * 1 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4
a4 = (i * 3 * n * tau) / (4 * ln[2]) + 1 + log_2(5) / 4

[@robertbriquet]

It's great but could skip some steps?

[@MAS1234P]

Using the definition of logarithm (rather than blindly applying log rules) will make the calculation faster.
4^2a = 80 =>2a = log80 (base 4). By definition

[@ahmedchabboubi1395]

Thank you

[@ChavoMysterio]

4ⁿ•4ⁿ=80
16ⁿ=80
2⁴ⁿ=2⁴•5
2^(4n-4)=5
4n-4=log_2(5)
4n=4+log_2(5)
n=1+¼[log_2(5)] ❤

[@angelnavarrete2300]

Excelente!!!

[@MasterTMathsClass]

Gracias! 😄

[@pierrettebalazut9407]

Sensas ! Tout devient facile avec vous

[@roguesoftware4free]

a=log_16(80)≈1.58

[@johneverson2433]

You must have taught math at my middle school, you didn’t explain anything that I could understand

[@AlexanderSemashkevich]

x=log80/2log4=(log16+log5)/4log2=1+log5/4log2
x≈1+0.699/(4×0.301)≈1903/2204≈1.5806
Deviation is about 0.0001
4^(2×1.5806)≈80.026

[@medium_codezone]

1.58

[@matzlwien]

4^a*4^a=80
4^a^2=sqrt80^2
4^a=sqrt80
Ln4*a=ln sqrt80
a= ln sqrt80/ln4
a= 1,580482
Thatś it, why do many steps???

[@MasterTMathsClass]

👌

[@aligundogdu5328]

a = (log5/log16) + 1 = 1,58...

[@MasterTMathsClass]

✌️

[@anniesworldeverythingsurro8214]

a =1.5805

[@aminjonimomnazarov167]

TThhaannkk yyoouu

[@MasterTMathsClass]

You're welcome

[@aminjonimomnazarov167]

@@MasterTMathsClass very nice