Japanese | Math Olympiad Logarithmic Equation | Exponent Simplification |
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- เผยแพร่เมื่อ 26 พ.ค. 2024
- Math Olympiad Logarithmic Equation | Exponent Simplification #exponent#olympiad #simplification#exam
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Math Olympiad Logarithmic Equation: (x - 5)^[log(5x - 25)] = 2; x = ?
(x - 5)^[log(5x - 25)] = (x - 5)^log[5(x - 5)] = (x - 5)^[log5 + log(x - 5)]
log(x - 5)^[log5 + log(x - 5)] = [log5 + log(x - 5)]log(x - 5) = log2
[log(x - 5)]² + log5[log(x - 5)] - log2 = 0
Let: y = log(x - 5), y² + (log5)y - log2 = y² + [log(10/2)]y - log2 = 0
y² + (log10 - log2)y - log2 = (y - log2)(y + log10) = 0
y - log2 = 0, y = log2 or y + log10 = 0, y = - log10 = log(1/10)
y = log(x - 5) = log2, x - 5 = 2; x = 7 or x - 5 = 1/10; x = 5 + 1/10 = 5.1
Answer check:
x = 7: (x - 5)^[log(5x - 25)] = 2^(log10) = 2¹ = 2; Confirmed
x = 5.1: (0.1)^(log0.5) = 2; Confirmed
The calculation was achieved on a smartphone with a standard calculator app
Final answer:
x = 7 or x = 5.1
Nice job !
Thanks!
Fantastic
Thank you so much 😀
Nice 2 & 5 equation... 💌
Keep watching
Nice job ❤
Thanks 😆 I'm glad you like it 👍
Amazing
Thanks!
Solved it in my head in 30 sec!
Can you solve it overhead for the second solution?
❤❤
Detailed explanation ! Good !
Glad it was helpful!
That`s great teacher.
Appriciated, that's why computer is invented
When you have this:
log(5m) * log(m) = log(2)
you can do this, 'cause log(10) = 1:
log(5m) * log(m) = log(2)^log(10)
then, you have
log(5m) * log(m) = log(10) * log(2)
log(5m) * log(m) = log(5 * 2) * log(2)
then m = 2, 'cause the expressions on both sides are the same
so:
x - 5 = 2
x = 7
Yes. And there's one more solution x=5.1
The other solution should more easier
5x-25 can't be 0
So 5x-25= 0
Here x= 5
Which means it should be 5.1 approximately,
multiplying both sides by 5, then let y=5x-25. we have (logy)^2=1, ....
Excelente, pero en el discriminante también se puede cambiar 2 = 10/5 y trabajar con log 5, entonces:
log2 = log(10/5)=log10-log5=1-log5
Great but results need verification, X can’t have two different acceptable values
Both are acceptable.
ln(x - 5)[ln(5x - 25)]/ln10 = ln2
ln(x - 5) = u
u(u + ln5) = (ln2)(ln2 + ln5)
u² + uln5 - (ln2)(ln2 + ln5) = 0
u = (-ln5 ± √[ln²5 + 4ln²2 + 4(ln2)ln5]/2
u = [-ln5 ± (ln5 + 2ln2)]/2
u = ln2 => ln(x - 5) = ln2 => *x = 7*
u = -ln10 => x - 5 = 1/10 => *x = 5.1*
the integer solution is x = 7.
Надо было сразу указать, что log -- это ln, то есть логарифм по основанию 10.
ln - это натуральный логарифм, по основанию е. А логарифм по основанию 10 пишется lg. Согласна, что тут непонятно в условии, что за логарифм, т.к. написано просто log без основания. Так что либо автор ошибся, либо это местная форма записи такая, мол, без основания это значит по 10. Но я сомневаюсь.
This is to challenge our brain and math concepts. Dependence on computers is for people who don’t think for themselves. It’s for zombies! Dr. K
AKA Democrats.
You could read "War and Peace" three times cover-to-cover and he would not have solved this problem .After logging both sides to base ten, work with ratios. Much easier!
А проще нельзя?
Логарифмировать по основанию 10
Получаем
log(x-5)* log(5*(x-5))=log2,
Есть формула
log (a*b) =loga+log b
Отсюда
log (x-5)*(log(x-5)+log5) =log2
И замена log (x-5)=a
Получаем обычое квадратное уравнение
а*(а+log 5)=log2
Тут фишка не в том, чтобы привести к квадратному уравнению, а в том, чтобы извлечь корень из дискриминанта.
Дискриминант
(log5)^2-4 log2
Если учесть что 10=5*2, а log 10=1 и log(a*b) =loga+logb, то
Получаем полный квадрат, как в решении и гораздо быстрее
i found 7 without pencil or page. but 5.1? oh no i can't do that lol
Same here
Why you dont write 10 in log index ?
The convention I use in the video to write log 10 is also acceptable
Or, you could just say let's try x=2 and see what happens.😄