Japanese | Math Olympiad Logarithmic Equation | Exponent Simplification |

[@walterwen2975]

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

[@Banana_Cat1011]

WE BE BECOMING GENIUS WITH THIS ONE🗣️🗣️🗣️🗣️🗣️🗣️🔥🔥🔥🔥🔥🔥🔥🐐🐐🐐🐐🐐🐐

[@UKPEINDANIELU.]

Fantastic

[@superacademy247]

Thank you so much 😀

[@KT-vi7gd]

Detailed explanation ! Good !

[@superacademy247]

Glad it was helpful!

[@kassuskassus6263]

Nice job !

[@superacademy247]

Thanks!

[@naakatube]

Solved it in my head in 30 sec!

[@superacademy247]

Can you solve it overhead for the second solution?

[@gelbkehlchen]

Solution: (x-5)^[lg(5x-25)] = 2 ⟹ (x-5)^{lg[5*(x-5)]} = 2 ⟹ (x-5)^[lg(5)+lg(x-5)] = 2 |lg() ⟹ [lg(5)+lg(x-5)]*lg(x-5) = lg(2) ⟹ lg(5)*lg(x-5)+[lg(x-5)]² = lg(2) |with u = lg(x-5) ⟹ lg(5)*u+u² = lg(2) |-lg(2) ⟹ u²+lg(5)*u-lg(2) = 0 |p-q-formula ⟹ u1/2 = -lg(5)/2±√{[lg(5)/2]²+lg(2)} = -lg(5)/2±√{[lg(5)]²/4+4*lg(2)/4} = -lg(10/2)/2±1/2*√{[lg(10/2)]²+4*lg(2)} = -[lg(10)-lg(2)]/2±1/2*√{[lg(10)-lg(2)]²+4*lg(2)} = -[1-lg(2)]/2±1/2*√{[1-lg(2)]²+4*lg(2)} = -1/2+lg(2)/2±1/2*√{1-2*lg(2)+lg²(2)+4*lg(2)} = -1/2+lg(2)/2±1/2*√{1+2*lg(2)+lg²(2)} = -1/2+lg(2)/2±1/2*√[1+lg(2)]² = -1/2+lg(2)/2±1/2*[1+lg(2)] ⟹ u1 = -1/2+lg(2)/2+1/2*[1+lg(2)] = -1/2+lg(2)/2+1/2+lg(2)/2 = lg(2) and u2 = -1/2+lg(2)/2-1/2*[1+lg(2)] = -1/2+lg(2)/2-1/2-lg(2)/2 = -1 ⟹ 1st case: lg(x1-5) = u1 = lg(2) |10^() ⟹ x1-5 = 2 |+5 ⟹ x1 = 2+5 = 7 2nd case: lg(x2-5) = u2 = -1 |10^() ⟹ x2-5 = 10^(-1) = 0,1 |+5 ⟹ x2 = 0,1+5 = 5,1

[@superacademy247]

Awesome 💯😎

[@carlosharmes2378]

Nice 2 & 5 equation... 💌

[@superacademy247]

Keep watching

[@RachidAburachid]

Nice job ❤

[@superacademy247]

Thanks 😆 I'm glad you like it 👍

[@rfnobando]

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

[@superacademy247]

Yes. And there's one more solution x=5.1

[@ethioturkish1821]

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,

[@kofimensahkarikari3944]

A marathon maths skills.

[@superacademy247]

Absolutely 💯🤗

[@joserubenalcarazmorinigo9540]

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

[@miffyLake]

That`s great teacher.

[@yifengxiao]

multiplying both sides by 5, then let y=5x-25. we have (logy)^2=1, ....

[@masoudghiaci5483]

❤❤

[@АндрейПергаев-з4н]

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

[@victorfildshtein]

Тут фишка не в том, чтобы привести к квадратному уравнению, а в том, чтобы извлечь корень из дискриминанта.

[@АндрейПергаев-з4н]

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

[@SidneiMV]

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*

[@maxk6856]

Why you dont write 10 in log index ?

[@superacademy247]

The convention I use in the video to write log 10 is also acceptable

[@ludmilak9396]

Надо было сразу указать, что log -- это ln, то есть логарифм по основанию 10.

[@mmdexperiments3828]

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.

[@bvenable78]

I can't seem to evaluate the check for x=5.1 Can someone help?

[@superacademy247]

It's a solution!

[@bvenable78]

@@superacademy247 Oh, yes; I'm not doubting it. I'm just struggling to evaluate the check. (It's a "me fail" thing.) I'm hoping someone out there will help walk me through it is all.

[@Lost_City007]

Sir, I'm here to help you. From your comment what I understand is... You want to evaluate (x-5)^[log 5(x-5)] = ? when x= 5.1 Ok, write these on paper for better understanding (x-5) = (5.1-5) = 0.1...you got this. Next we've to find, (0.1)^log(0.5) Notice that , log(0.5) =log(1/2) =( log1 -log2) =( 0 - log2) = -log2....... We can write, (0.1) = (1/10) Therefore the whole thing becomes, (1/10)^(-log2) = 10^log2 = 2....... That's it 😉 Sir, if you still have any doubt reply me.

[@Lost_City007]

@@bvenable78 Sir, I'm here to help you. From your comment what I understand is... You want to evaluate (x-5)^[log 5(x-5)] = ? when x= 5.1 Ok, write these on paper for better understanding. (x-5) = (5.1-5) = 0.1...you got this. Next we've to find, (0.1)^log(0.5) Notice that , log(0.5) =log(1/2) =( log1 -log2) =( 0 - log2) = -log2....... We can write, (0.1) = (1/10) Therefore the whole thing becomes, (1/10)^(-log2) = 10^log2 = 2....... That's it 😉 Sir, if you still have any doubt reply me.

[@bvenable78]

@@Lost_City007 OMG thank you. I turned log(1/2) into log(1) + log(2) (instead of subtraction). Embarrassing. Thanks again for setting me right. :)

[@CawasKapadia]

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

[@kevinmadden1645]

AKA Democrats.

[@kevinmadden1645]

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!

[@tristan583]

Great but results need verification, X can’t have two different acceptable values

[@superacademy247]

Both are acceptable.

[@avengersage4354]

But x=10 is satisfying the question

[@avengersage4354]

But x=10 is satisfying the upper questions so can x=10 be it's solution 🤔

[@superacademy247]

Yes. It's a solution!

[@avengersage4354]

@@superacademy247 thanks a lot sir

[@sunil.shegaonkar1]

the integer solution is x = 7.

[@ardayavuz9806]

i found 7 without pencil or page. but 5.1? oh no i can't do that lol

[@sirajzama8080]

Same here

[@umutkargili3617]

Appriciated, that's why computer is invented

[@cookiemonster-nk3xb]

Shut your pie hole

[@baselinesweb]

Or, you could just say let's try x=2 and see what happens.😄

[@hossamwalid3599]

Amazing

[@superacademy247]

Thanks!