A Quick and Easy Problem on Exponential Expressions

[@bprpfast]

😮!! I would have just done the log. This is a very nice solution!

[@SyberMath]

Thank you!!! 💖

[@diptirindani8190]

Me too

[@manojsurya1005]

I took log on both sides and then 1/a+1/b+1/c=log_42 (2)+ log_42 (3)+ log_42 (7) =1, Ur method without log is amazing😉

[@prathikkannan3324]

Used same method

[@SyberMath]

Nice! Thank you! ☺️

[@advaitpetiwale9596]

The moment you got 2^1=42^(1/a), I realized what you were doing and said "whoa, no no no no. That was waaay clever". And really, it was.

[@SyberMath]

💖

[@vishalmishra3046]

*Taking log on both sides, a log(2) = b log(3) = c log(7) = log(42) Therefore. 1/a + 1/b + 1/c = log(2)/log(42) + log(3)/log(42) + log(7)/log(42) = log(2x3x7)/log(42) = log(42)/log(42) = 1*

[@MathElite]

I first solved this with taking log on both sides but now i realize prime factorization solves this nice problem, Syber

[@SyberMath]

Thank you, Math Elite!

[@MyOneFiftiethOfADollar]

Change of base to ln and straightforward algebra leads to ln(42)/ln(42)=1 The way it was presented was neat method where properties of logarithms were not needed

[@echandler]

Nice problem. Like others, I took logs. Your solution provided an “aha” moment that was quite satisfying.

[@SyberMath]

Great to hear!

[@tbg-brawlstars]

I am in class 10 and even don't know what are logarithms but you taught me that we can do questions just by basics and using our brain Thank You :))

[@dolos_kuroshoe]

Okay. I figured out the trick. I hope someone else does too before the video starts. I’m not spoiling anything.

[@SyberMath]

😁

[@MathNerd1729]

Don't worry; you're not the only one! I saw the trick too! Good job! :)

[@damianbla4469]

It was given that: 2^a = 3^b = 7^c = 42 That means that at the same time: 1) 2^a = 42 2) 3^b = 42 3) 7^c = 42 1) a = log.2(42) 2) b = log.3(42) 3) c = log.7(42) Using the formula "1 / log.a(b) = log.b(a)" we get: 1) (1/a) = 1 / log.2(42) = log.42(2) 2) (1/b) = 1 / log.3(42) = log.42(3) 3) (1/c) = 1 / log.7(42) = log.42(7) (1/a) + (1/b) + (1/c) = = log.42(2) + log.42(3) + log.42(7) = Using the formula "log.a(b) + log.a(c) + log.a(d) = log.a(b * c * d)" we get: = log.42(2 * 3 * 7) = = log.42(42) = = 1 Answer: (1/a) + (1/b) + (1/c) = 1

[@kuppuswamyssundaram7350]

Solution without Logarithm 2^a = 42 implies 2 = 42 ^ (1/a) Similarly 3 = 42^ (1/b) & 7 = 42 ^ (1/c) 2*3*7 = 42 {42^(1/a)} {42^(1/b) }{42^(1/c)} = 42 42 ^ { (1/a)+(1/b)+(1/c)} = 42 ^1 So (1/a)+(1/b)+(1/c) = 1

[@sandanadurair5862]

Good teacher indeed!!!!!

[@HuzaifaMuhammadSIddique]

I thought there was no other way than just doing log. But you proved it. Well done man 👍

[@SyberMath]

Thanks 👍

[@tonyhaddad1394]

Wowww what an intersting answer (1) i solved by changing the log base to a unique one ( i chose 2) , then used the log propertie ......done!

[@quantumobject3815]

I love how simple yet beautiful this is 🤗

[@SyberMath]

Thank you! 😊

[@denismilic1878]

Finally, this is the first your video that I solved the same way as you.

[@bharathkarkera3223]

Man @SyberMath This is some excellent mathematical skill.. appreciate it

[@roberttelarket4934]

Another beautiful and excellent problem and solution! Keep them coming!

[@SyberMath]

Thanks, will do!

[@adandap]

But the log approach is easier (at least I think it is, though the methods are equivalent). Take all logs to the same base (doesn't matter which). a = log(42)/log(2), b=log(42)/log(3), c = log(42)/log(7). Then 1/a + 1/b + 1/c = (log(2)+log(3)+log(7))/log(42) = log (2.3.7)/log(42) = log(42)/log(42) = 1.

[@falknfurter]

If you are not comfortable with log's this problem might be a first step to get on friendly terms with them!

[@呂永志]

0:55 If students have learned log, this method is even better. But method 2 can solve the question without using log.

[@SyberMath]

That’s what I like about it. That makes the problem more accessible. It also uses a neat trick.

[@swenji9113]

Wait what? How could students know what is a^b without knowing the log? I believe the usual definition of a^b involves the log no matter where you learn maths, doesn't it?

[@UECAshutoshKumar]

This is a really nice approach ❤️

[@SyberMath]

Thank you!!! 💖

[@kailashanand5086]

yeah i solved it using the second method, mostly because im preparing for olympiads haha another great question, with a beautiful solution :D

[@SyberMath]

Fantastic!

[@heeraparesh3558]

Wow its such a great method! So helpful

[@SyberMath]

Thank you!

[@242math]

very well done, thanks for sharing Syber

[@SyberMath]

My pleasure!

[@jnx6558]

Bonjour, on trouve aussi de la façon suivante (bien que votre méthode soit plus intéressante): a = ln42/ln2, b= ... etc. du coup, au final, on a: (ln2+ln3+ln7)/ln42 = 1

[@prabhudasmandal6429]

Interesting problem, nice solution.thanks.

[@SyberMath]

You're welcome

[@scottleung9587]

I solved it using logarithms, but your solution is brilliant!

[@SyberMath]

Thank you! 😊

[@darkavenger6407]

I learned the coolest way back in school that you just showed. The fact that this can also be solved using log astonished me a bit😂😅

[@SyberMath]

😁

[@-basicmaths862]

Easy and intresting a log 2basek=1, k=42 log 2basek=1/a log3 base k=1/b log 7basek= 1/a+1/b+1/c= log2basek+log3basek+log7basek 1/a+1/b+1/c=log42basek=1

[@alainrogez8485]

Brillant. I used log but I was stuck.

[@SyberMath]

😊

[@vuyyurisatyasrinivasarao3140]

Good solution sir

[@nicogehren6566]

nice approach sir thanks

[@SyberMath]

Most welcome

[@chandrashekharmehta6121]

Wow!! Good job syber

[@SyberMath]

Thank you! ☺️

[@swátt7ik]

I like when you use logarithms and that make the video super easy ✌✌🙏🙏🙏🙏😊😊😊

[@mohammadroshani936]

Very very thank. This well....

[@golddddus]

Finally - Problem with Number 42. The most important Number in the Universe! 🤗

[@robertturpie1463]

Very neat

[@user-dz6pi4sm8u]

Very tricky problem.

[@omaralvarezzaleta4728]

Excelente,para olimpiadas de matematicas

[@SyberMath]

Thank you!

[@KJ-zs7pi]

Oh man what a great soln loved it Just doesn't strike this method ,huh?

[@77Chester77]

Wow, super cool technique

[@SyberMath]

Thank you!

[@citylights940]

Now i slove this pobleam thanks bro

[@SyberMath]

No problem

[@yoav613]

This video shows us that even the ugliest numbers can be combined to nice number

[@SyberMath]

That’s the beauty of math! 🤩

[@mathsfamily6766]

That’s so great , I love the you show so much 😍🥰

[@SyberMath]

Thank you so much 🤗

[@AnuragSingh_1729]

Given that 2^a=3^b=7^c=42 2=(42)^1/a ---------(1) 3=(42)^1/b ----------(2) 7=(42)^1/c ------------(3) (42)^(1/a)×(42)^(1/b)×(42)^1/c= 2×3×7 (42)^{1/a+1/b+1/c} =42 On comparing power both side we get 1/a+1/b+1/c=1 Which is correct answer 👍

[@DrWeselcouch]

How do you come up with so many great questions???

[@aashsyed1277]

he just does! by the way, if you wanna ask questons, then i reccommend commenting earlier.

[@SyberMath]

Thanks for the kind words! Great question! Problems come from many different resources. I have been writing and solving problems for quite some time, something I enjoy to do! 🤩

[@amirhassanpour3730]

Very clever!!!

[@SyberMath]

Thank you! 😊

[@himejipyonkichi7492]

Hello from Japan😀 interesting！！！

[@SyberMath]

Hello 😊

[@luzherrera4463]

Is there a book that teaches to solve this kind of "a,b,c" problems? I've seen them everywhere in math olympiads but never found a book dealing with them.

[@muhammadhussainsarhandi9928]

So why don't you write and publish your own book to solve this type of problems?

[@pietergeerkens6324]

Very nice!

[@SyberMath]

😊

[@1Rom4ik]

I used switching the base of logarithms, which gave me the same answer)

[@SyberMath]

Very good!

[@inboxingleads6336]

If this would've been a multiple choice question, and it figured 0 or 1 or infinity or √42 i.e. something relatable, I would have gone for that 😀.

[@chessdev5320]

It was very easyyy

[@SyberMath]

Ok

[@xy4611]

I solved this on my mind with logarithm and without pencil i feel clever lol

[@kuppuswamyssundaram7350]

Nice problem

[@SyberMath]

Thanks

[@aashsyed1277]

wow thanks so much sybermath! hope you are fine :DD

[@mathsislove197]

Hi

[@aashsyed1277]

@@mathsislove197 how are u?

[@mathsislove197]

Fine what about you?@@aashsyed1277

[@aashsyed1277]

@@mathsislove197 i am fine what were you doing?

[@rssl5500]

Nice problem :D

[@SyberMath]

Thank you!

[@nikitakipriyanov7260]

Intuition said: one. Checked: one.

[@Yuuki3uwu]

The second method is more elegant indeed

[@aashsyed1277]

This is so nice!!!!!!!

[@SyberMath]

Thank you! 💖

[@DrWeselcouch]

hello......

[@aashsyed1277]

@@DrWeselcouch hi! good to see ya

[@aashsyed1277]

@@SyberMath check out @Dr Wesselcouch's channel!

[@SyberMath]

@@aashsyed1277 I did and it's really cool!

[@yhamainjohn4157]

So nice ...

[@SyberMath]

Many many thanks

[@ufukkeskin.]

Nice

[@SyberMath]

Thanks

[@poseidon.M]

Perfect💖👍🍻

[@shatishankaryadav8428]

2×3×7 have same power then if we write 1/a +1/b+1/c =1 easy solutions

[@victoralvarez3519]

I wonder if it is possible to find the values of a,b and c

[@Neco3376]

By definition a is the log_2(42) and similar for b, c so those are the exact values.

[@timetraveller2818]

yes ,. it is because 2^a=3^b=7^b=42

[@grosman4221]

Wow, it was suddenly

[@3r3nite98]

Imma solve this myself,also is it that bad to call a easy problem (that seems difficult to the person) hard. Be cos Drama happened once previously. Nevermind I didn't solve it myself.

[@SyberMath]

You can call it easy or hard. It’s relative. No one should be offended by that because if you say easy, it means easy for you but if you say oh this is so easy, I can do it with my eyes closed or I can do it in 5 seconds and keep bragging about it or argue with other people, then it is not nice.

[@3r3nite98]

@@SyberMath OK thx. Also at the video where I commented "this sure is easy",I was being sarcastic. Also the reason why I commented this is,be cos of adway kumar (I'm not saying that they're a bad person,but I wonder if they're older than me,I mean I don't want to complain,but they get prioritized over me most of the time,but I'm trying to get along with them,also I regreted saying my age,I mean you didn't heart my comment so I think you think that all x year olds,where x is my age are toxic).

[@cuongtu6088]

good

[@Spectre021]

It's a simple spell but quite unbreakable.

[@SyberMath]

😁

[@rangaswamyks8287]

Very easy i got amswer In my mind... Didnt worked on paper

[@susennath6035]

Know the presentation first. You have so much bookish knowledge

[@SyberMath]

😂🙃😂

[@АзиретАкматбеков-й1м]

Hah, nice question!

[@HuzaifaMuhammadSIddique]

But this method won't work all the time since you won't get the bases same always. This was a exceptional case.

[@SyberMath]

That is true

[@anyaleny1258]

What program do you use?

[@SyberMath]

Notability

[@anyaleny1258]

@@SyberMath Thanks.

[@SyberMath]

@@anyaleny1258 No problem!

[@enejidjsi5939]

1:10 this is just from my perspective, but is it possible you could skip obvious steps? I mean I think everyone watching this would know basic properties of logarithms, so why not just directly write a = log_2 (42)?

[@Wilson3.1415]

I hope you can provide Chinese subtitles,please.......🥺 Your videos is so nice, but I can't understand English very well...😔

[@SyberMath]

Sure

[@aronjorg5921]

I don't think that the logic is correct so please help me: Factoring 42=2*3*7 which means 2'3*3'3*7'3=2'a*3'b*7'c. In this equation we have solutions for a=b=c=3 (however it isn't satisfy the original), but notice that the result is correct 1/a+1/b+1/c=1. Why?

[@Metal_dead]

You can do the same with logarithms

[@dennisphilip7596]

By taking log , we can solve it

[@victorchoripapa2232]

It's one (I didn't see the video)... Am I right?

[@SyberMath]

You can check 🙃😉

[@RosinaldoSSouza]

Genial! Rsrsrs

[@theartisticactuary]

a=ln42/ln2, b=ln42/ln3, c=ln42/ln7 Take reciprocals and add them up to get (ln2 + ln3 + ln7) / ln42 = ln42/ln42 So much quicker

[@SyberMath]

Cool!

[@quockhainguyen4704]

simple

[@SyberMath]

Ok

[@bandarisunny4455]

Its easy one

[@adhiraiselvanengineer]

I didn't use log or Ln at all

[@SyberMath]

Good!!!

[@rssl5500]

Sadly I didn’t solve this

[@3r3nite98]

Me neither.

[@Сайд-магамедАбдулхамидов]

I like it’s?

[@Germankacyhay]

👍❤

[@SyberMath]

💖

[@rssl5500]

Don’t know logarithms btw lol I know basics from internet and KZbin but nothing high lvl lol

[@SyberMath]

Logs have their own rules. Understanding the definition is important. You can turn a logarithmic expression into an exponential one using the definition.

[@mathsislove197]

This is not high level tho. What’s your age?

[@rssl5500]

Im 14 and this year I’m going to 9th grade in here we learn logs and 10-12th grade I think

[@mathsislove197]

oh @@rssl5500

[@synergetic4d]

Safe?? No! Stay free! Not safe

[@rssl5500]

Wow it’s 1 :D

[@Qermaq]

Solved it the hard way and then I saw this way. Oops!

[@giuseppemalaguti435]

1

[@mathevengers1131]

Can you prove (199!!)*(-201!!) = 1 Where x!! = x(x-2)(x-4)...

[@ThAlEdison]

No, the way I'm reading things x!! doesn't end, so you'd just end up with it always approaching negative infinity. If you say it ends at positive 1, then you have -201*(199!!)^2=1, and I don't think 199!!=isqrt(201) Maybe if it's (-201)!!, but then you have the negative infinity thing again. There may be some property you're trying to show, but I don't understand it?

[@mathevengers1131]

@@ThAlEdison I have a made a video about it on my channel. You will fully understand it there.

[@mathevengers1131]

@@ThAlEdison factorial of any negative integer is infinity or -infinity but double factorial of any odd negative integer ( For even negative integers it diverges ) converges. So it has some value. It may be positive or negative. Just see my yesterday's shorts you will understand what I am trying to say.

[@pardeepgarg2640]

" *Why to pass a math exam if you can pass away* " Mestien, Living in Newtown

[@neutron417]

Oral Question

[@gachanguyen7154]

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[@sssh024]

0

[@sssh024]

1,typed too quick

[@footballlife5870]

👍👍👍

[@일상남선생]

1