Germany | Can you solve this? | Math Olympiad

[@maths01n]

Liked your consistency ❤

[@MasterTMathsClass]

Thank you 🙌

[@samuelkelly2197]

One can never go wrong with Master T Maths Class. He makes Indices and Logarithms questions and solutions a child’s play. A very good Tutor. BRAVO once again Sir

[@MasterTMathsClass]

Thank you very much 💙

[@stpat7614]

There's a complex root too:
2^x * 2^x = 20
(2^x)^2 = 20
sqrt([2^x]^2) = +/- sqrt(20)
sqrt([2^x]^2) = +/- sqrt(2^2 * 5)
sqrt([2^x]^2) = +/- sqrt(2^2) * sqrt(5)
([2^x]^2)^(1 / 2) = +/- (2^2)^(1 / 2) * 5^(1 / 2)
(2^x)^(2 * [1 / 2]) = +/- 2^(2 * [1 / 2]) * 5^(1 / 2)
(2^x)^1 = +/- 2^1 * 5^(1 / 2)
2^x = +/- 2 * 5^(1 / 2)
(1 / 2) * 2^x = +/- (1 / 2) * 2 * 5^(1 / 2)
2^x / 2 = +/- 1 * 5^(1 / 2)
2^x * 2^(-1) = +/- 5^(1 / 2)
2^(x - 1) = +/- 5^(1 / 2)
2^(x - 1) = + 5^(1 / 2), or 2^(x - 1) = - 5^(1 / 2)
Suppose 2^(x - 1) = + 5^(1 / 2)
2^(x - 1) = 5^(1 / 2)
log(2^[x - 1]) = log(5^[1 / 2])
(x - 1) * log(2) = (1 / 2) * log(5)
(x - 1) * log(2) / log(2) = (1 / 2) * log(5) / log(2)
(x - 1) * 1 = log(5) / (2 * log[2])
x - 1 = log_2(5) / 2
x - 1 + 1 = log_2(5) / 2 + 1
x = log_2(5) / 2 + 1
x1 = log_2(5) / 2 + 1
Suppose 2^(x - 1) = - 5^(1 / 2)
2^(x - 1) = - 5^(1 / 2)
2^(x - 1) = -1 * 5^(1 / 2)
ln(2^[x - 1]) = ln(-1 * 5^[1 / 2])
(x - 1) * ln(2) = ln(-1 * 5^[1 / 2])
(x - 1) * ln(2) / ln(2) = ln(-1 * 5^[1 / 2]) / ln(2)
(x - 1) * log_2(2) = ln(-1) / ln(2) + ln(5^[1 / 2]) / ln(2)
(x - 1) * 1 = ln(e^[i * tau / 2]) / ln(2) + (1 / 2) * ln(5) / ln(2)
x - 1 = (i * tau / 2) * ln(e) / ln(2) + ln(5) / (2 * ln[2])
x - 1 = (i * tau / 2) * 1 / ln(2) + log_2(5) / 2
x - 1 = i * tau / (2 * ln[2]) + log_2(5) / 2
x - 1 + 1 = i * tau / (2 * ln[2]) + log_2(5) / 2 + 1
x = i * tau / (2 * ln[2]) + log_2(5) / 2 + 1
x2 = i * tau / (2 * ln[2]) + log_2(5) / 2 + 1
{x1, x2} = { [log_2(5) / 2 + 1], [i * tau / (2 * ln[2]) + log_2(5) / 2 + 1] }

[@NumberPhi]

Beautiful and didact resolution!
Thank you and Happy New Year!

[@MasterTMathsClass]

🙏 Thanks!

[@hulyaaycan4660]

Teşekkürler ve Mutlu Yıllar!

[@dahupaylo]

2^2x= 5x2^2.
Then take log of both sides. Much simpler.

[@jozsefsalagvardi7694]

2^x*2^x=20
2^2x = 2*10 : log
2x*lg(2) = lg(2) + lg(10)
2x*0.3010=0.3010 + 1
0.6020 X = 1.3010
x= 1.3010/0.6020 = 2.1611
(2^2.1611)^2=(4.4725)^2 >20.003

[@МихаилИванов-х7ч]

Всё понятно. Просто у меня один вопрос по какому основанию он изначально логорифмирует? Это не критика мне просто не понятно.

[@Diph64]

По основанию 10. То, что у них называется log, у нас называется lg

[@yanssala2214]

X = 2,1609....

[@AlexanderSemashkevich]

x=(log4+log5)/2log2=1+log5/2log2
x≈1+0.699/(2×0.301)≈1301/602≈2.161
Deviation is about 0.0002
2^(2×2.161)≈20.001

[@MasterTMathsClass]

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