Australian Intermediate Mathematics Olympiad.Preliminary Lambert w function. Exponential Equation.

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[@dianedong1062]

I've been struggling with this type of problem for a long time, I'm finally starting to get it.

[@thebasisti2482]

Wow

[@dianedong1062]

Thank you for clearly going through this step by step.

[@superacademy247]

Glad it was helpful!

[@Valhondrian]

Mathematics should be about minimizing the time and complexity to give an answer on a given question. Using W function is overkill to answer this very simple question. 1) Observe that the equation has a trivial solution x=5 because 2^5+5=32+5=37 2) Let f(x) = 2^x+x-37 and we look for x such as f(x)=0 f'(x)=ln2.2^x+1 >0 =>f(x) is strictly increasing over |R . Moreover f(0)=-350 when x goes to + infinity => there is a unique x in ]0,+infinity[ such as f(x)=0. But as we know that x=5 is solution, it is the unique solution of the equation f(x)=0. QED

[@bouzekriamir1539]

I am a 64 year old, and i love math and i am a butcher and when i vind the time i see the math problème, tanks ser

[@superacademy247]

You're welcome. Thanks 👍 💯 😊 for your support

[@bjorncedervall5291]

Wow! Saw the answer directly but didn't know how to logically solve it. Never imagined how complicated (and elegant) the solution could be. Did this type of stuff in school more than 50 years ago - would probably have been a tough challenge then too. Now follow these types of challenges as entertainment and to keep my brain office in some type of shape.

[@superacademy247]

I'm pleased that you found this to be of assistance.

[@bjorncedervall5291]

@@superacademy247 Smile & thumb up!

[@superacademy247]

@@bjorncedervall5291 Thanks!

[@simonpayne7994]

Really wonhderful"

[@slavinojunepri7648]

Excellent use of the Lambert function.

[@superacademy247]

Thanks for your nice observation

[@superacademy247]

Thanks for your nice observation

[@lechaiku]

If you can use 2^5=32 at 6:54 You can use the same at the begining. 2^x + x = 37 2^x + x = 32 + 5 2^x + x = 2^5 + 5 Since the functions 2^x = 37 -x has 1 unique solution we can easily compare both sides (the same base and the same operation (adding) 2^x = 2^5 x = 5

[@zakzakzak3345]

Только посмотрел на пример и через секунду уже знал ответ... I just looked at the example and in a second I already knew the answer ....

[@superacademy247]

Glad to hear that

[@RobertHayes-gi6dz]

thank you

[@superacademy247]

Welcome!

[@renesperb]

The solution for the general case a^x=bx+c is - (1/ln a* W[ - a^(-c/b)*l(n a)/b ] +c/b ).For a second solution one has to replace W[x] by W[-1,x] .

[@wojtekostrowski456]

How about the derivative of LHS is positive, hence it's monotonically increasing hence only one real solution x=5?

[@superacademy247]

Awesome 💯

[@robertgapatas]

Melbourne-Albury

[@heniwatisetiono6995]

x=5

[@josejefferson2812]

Using common sense it is possible to assume rhe value of x as 5.2™5=32,32+5=37.

[@jim2376]

On inspection, 32 + 5 = 37. Who needs Lambert?

[@superacademy247]

Examiner

[@jim2376]

@@superacademy247My last math class was in 2009. Examiners are irrelevant to me.

[@existing666]

Nice!! I like w lambert and find it so fun to dee the algebra puzzle come together

[@superacademy247]

Thank you! Cheers!

[@existing666]

Just wondering, can we describe imaginary solutions? I'm wondering if anywhere where there is a w Lambert function we could take countably infinitely many different branches. Is there a clean way to describe any of those solutions if they are valid?

[@superacademy247]

Yes, there's a practical way of doing it. Graphical method to explain the inverse of a function.