Germany | Can you solve this? | Math Olympiad Problem

[@serjonunes181]

2xlog 2 = log 100 x = 2/2log2 x = 1/log2

[@nestor6565]

This is the most direct, quick and smart solution. I guess that this video went last at math olympiad

[@robertoquiroz2782]

Justo así pensé resolver la ecuación 😊

[@BrijLalSahu-bb1lu]

🎉🎉 very nice @MATHS TUTORIAL ( BL SAHU)

[@stpat7614]

Including the complex root: 4^x = 100 4^x / 4 = 100 / 4 4^(x - 1) = 25 (2^2)^(x - 1) = 25 2^(2 * [x - 1]) = 25 2^([x - 1] * 2) = 25 (2^[x - 1])^2 = 5^2 sqrt([2^(x - 1)]^2) = +/- sqrt(5^2) 2^(x - 1) = +/- 5 ln(2^[x - 1]) = ln(+/- 5) (x - 1) * ln(2) = ln(+/- 5) (x - 1) * ln(2) = ln(5), or (x - 1) * ln(2) = ln(-5) Suppose (x - 1) * ln(2) = ln(5) (x - 1) * ln(2) = ln(5) (x - 1) * ln(2) / ln(2) = ln(5) / ln(2) (x - 1) * log_2(2) = log_2(5) (x - 1) * 1 = log_2(5) x - 1 + 1 = log_2(5) + 1 x = log_2(5) + 1 Suppose (x - 1) * ln(2) = ln(-5) (x - 1) * ln(2) = ln(-5) (x - 1) * ln(2) = ln(-1 * 5) (x - 1) * ln(2) = ln(i^2 * 5) (x - 1) * ln(2) / ln(2) = ln(i^2 * 5) / ln(2) (x - 1) * log_2(2) = ln(i^2) / ln(2) + ln(5) / ln(2) (x - 1) * 1 = ln(e^[i * tau / 2]) / ln(2) + ln(5) / ln(2) x - 1 = (i * tau / 2) * ln(e) / ln(2) + ln(5) / ln(2) x - 1 = (i * tau / 2) * 1 / ln(2) + ln(5) / ln(2) x - 1 = i * tau / (2 * ln[2]) + log_2(5) x - 1 + 1 = i * tau / (2 * ln[2]) + log_2(5) + 1 x = i * tau / (2 * ln[2]) + log_2(5) + 1 x1 = i * 0 * tau / (2 * ln[2]) + log_2(5) + 1 x2 = i * 1 * tau / (2 * ln[2]) + log_2(5) + 1

[@TheMathManProfundities]

4ˣ = 100 ⟺ x = log₄100 = 1 + log₄25 = 1 + 2log₄5

[@StevenTorrey]

Sure enough: log 5 (0.6989) / log 2 (0,3010) = 2.3219 + 1 = 3.219; 4^3.3219 = 99.996 = 100

[@Marquinhos_Aguiar]

👏👏👏

[@paolomarazzi7433]

(2^x)^2=10^2 2^x=10 x=3,32

[@xy-pk8gb]

Yes. (Just answering your question.)

[@schcl]

Why make it complicate? Log 100 is 2, no need to transform in log (4x25) etc!

[@someonespadre]

X=Log100/Log4≈3.32

[@brunozmuda9066]

trop compliquée ta solution. Log 4^x = log100 >>> x(log2+log2)= log10+log10 >> 2xlog2= 2 >>> X= 1/log2

[@rafaelyorro4466]

X=logbase2 of 5

[@Budgeman83030]

I stopped at 1 + log base 4 of 25

[@rafaelyorro4466]

X=3.33

[@PawełDąbrowski-n3w]

x = log100/log4 X = log25 + log4 /log4; X = 1+log_4_(25) X≈ 3,322

[@MasterTMathsClass]

✌️

[@AlexanderSemashkevich]

x=log100/log4=2/2log2=1/log2 x≈1/0.30103≈3.32193 4^3.32193≈100.00026

[@MasterTMathsClass]

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