Good job. The biggest takeaway for the kids is to be systematic and keep using the techniques and rules they have learned. Practice. Practice. Practice. Thanks.
77 в ст х -121 в ст х нужно было при извлечении из корня взять по молулю, т к выражение меньше 0. Еще замечание-просьба- очень мелко пишите. 😊 спасибо за работу!
Another solution is to divide both sides by 11^x. If ((7/11)^x -1)=t then the the equation is "t=sqrt(t)" => t is equal either 0 or 1. => x(1)=0; x(2)=lg2/lg(7/11)=lg2/(lg7-lg11)
the two solutions are both acceptable, even from the condtions of existance, after all it is necessary that the argument inside the root must be positive or null, and this is possible only if x
Wow!!! We are glad to hear this sir. This is our major objective and we are happy it is taking effect in the lives of some of our viewers. We pledge to do more by His grace so long it is within our ability as far as the channel is concern. So, keep watching and we will not disappoint you by His grace. Much love sir.💕💕💕💖💖❤️❤️
Noted, but OnlineMathsTV will appreciate it and will be forever grateful to you if and only if you can drop your approach to this challenge sir. Because, @onlinemathsTV we are 100% open to criticisms and corrections for we believe that is the sure way to learn new skills, unlearn outdated skills and to grew. Many thanks and much love @Pearlwood Sword.
Nice procedure @Jakes...👍👍👍
Many thanks to you sir.
Quite.elaborate
thank you, very well explained
This is really cul, quite different. Nice 👍 work dude and thanks for this approach tutor Jakes.
My pleasure!
Wow!!! This is great my brother, the sky is your starting point.
thanks ma, much love.
Good job. The biggest takeaway for the kids is to be systematic and keep using the techniques and rules they have learned. Practice. Practice. Practice. Thanks.
Thanks a million sir. Your words of encouragement are well received.
Much love❤️❤️💕💕💕
Nice work. I solved it using a different approach within a short period of time and still got same results.
Excellent!....👍👍👍you are the math guru sir.
Learning this trick is great. Thanks to onlinemathstv once again.
Thanks for leaving a new skill from our video.
I love master Jakes ...♥️♥️♥️
Smiles....thanks for loving me. I love you in turn also...💖😍😍
❤❤❤❤❤
Nice step, but working with log base at first is better than the base 7/11 you introduced.
Yes, you are right. Thanks for that contribution sir.
You have made calculation unnecessarily prolonged and critical
wow
Thanks a million sir.
Хорошо 🙂
nice question
Thanks sir.
Ótimo
Thanks for encouraging us sir.
We love you❤️❤️💖💖💕💕
thank you - very helpful
We glad you gained some values from this video tutorial sir.
Thanks for dropping a comment and we love you so much...❤️❤️💖💖
Didn't check your second result, but doesn't what's under the radical need to be greater than or equal to zero ?
77 в ст х -121 в ст х нужно было при извлечении из корня взять по молулю, т к выражение меньше 0. Еще замечание-просьба- очень мелко пишите. 😊 спасибо за работу!
Good explaination. But you can reduce the writing steps.
Another solution is to divide both sides by 11^x. If ((7/11)^x -1)=t then the the equation is "t=sqrt(t)" => t is equal either 0 or 1. => x(1)=0; x(2)=lg2/lg(7/11)=lg2/(lg7-lg11)
We can reduce the steps like this: a^x = m, then logm to base a = x instead of take logs on both sides.
the two solutions are both acceptable, even from the condtions of existance, after all it is necessary that the argument inside the root must be positive or null, and this is possible only if x
Wonderful👍👍👍
Many thanks to you sir.
@@onlineMathsTV thanks to you for this :)
😮
c'est bien , mais un peu long à mon gout : Continuez , vous serez plus rapide :)
Noted sir, I will try to increase my speed and thank you for the advice sir.
Much love❤️❤️💖💖💖💕💕
wow!
Bro, you've made me love maths
Wow!!! We are glad to hear this sir.
This is our major objective and we are happy it is taking effect in the lives of some of our viewers.
We pledge to do more by His grace so long it is within our ability as far as the channel is concern. So, keep watching and we will not disappoint you by His grace.
Much love sir.💕💕💕💖💖❤️❤️
Take a short solution for question because it will be easy
Noted, thanks for this wonderful suggestion sir.
We love deeply for this sir...💕💕💕
There's a mistake after squaring both sides. By law of indices 11 to power x multiplying by itself give you 11 raised to power x square.
How can I do away with sub titles in video
At the top right hand corner of your screen you will see "CC" touch it and it will stop showing subtitles.
Thanks sir.
☺вау
But 1= e^bi e, also
Nice solution but you are solving very slowly, it bores, can you please increase speed while you are solving
Можете ускорить видео в настройках над видео
lasolution:x=3;y=5
et
x=1;y=3
Comment trouverl le 2ème resultat?Merci
Is there a mistake with y square?
Not too clear with your comment/question sir.
Kindly rephrase it for a better understanding sir.
Thanks.
[√(7^x-11^x)]^2 = |7^x -11^x|
Merci
(SVP)
je veux résolu2^x-_3^y=5
Alright. Thanks for dropping this, the video will come out in no distant time.
squaring both sides isnt the best solution bcs its only implication
Noted, but OnlineMathsTV will appreciate it and will be forever grateful to you if and only if you can drop your approach to this challenge sir. Because, @onlinemathsTV we are 100% open to criticisms and corrections for we believe that is the sure way to learn new skills, unlearn outdated skills and to grew.
Many thanks and much love @Pearlwood Sword.
Comment trouver x=5ety=3
32_27=5????
Not too clear with your question sir.
Can you kindly rephrase it please?
@@onlineMathsTV Bonjour
je dis 2^x_3^y=5
solution
x=5;y=3
Comment trouver le deux résultats?
Merci
X=0
Bravo!!! You the master here
Is that necessary to set 77^x - 121^x >=0?
No sir, we cannot do away with the RHS in solving this challenge, besides that will end up giving us just one solution ignoring the second.
Why the condition would forget a solution?