i found the value of x by testing x = 2; you not only showed exact way how to find x=2, but also the second complex value of x by making clever use of Euler equation, thank you
@@onlineMathsTV Hi, i found how to get 2 of the equation: first, 9^x=3^2x, then 3^2x - 1 = (3^x+1)*(3^x-1) for a^2-b^2=(a+b)*(a-b), then 3^x+1 from up and 3^x+1 from down are gone, then 3^x-1=8, 3^x=9, x=2
I am a Taiwanese, watching this random video at my 3AM😂While I would go to work at 9AM😂Why am I here How about make 9^x-1 equals to (3^x+1)*(3^x-1), and then literally divide 3^x+1, we got 3^x-1=8, thus x=2; Considering 3^x+1=0 for a expectation, thus x=blablabla
Using the difference of two square for the numerator as you have rightly stated will only give the real value x=2 thereby omitting the imaginary root. In case of exams where you are asked to solve for only the real root then this approach is the fastest and the best. Thanks for your contribution and thanks for creating out time to watch our contents despite your tight schedule. We love you sir.
@@onlineMathsTV Hi, I would like to thank you for throughly replying, and also for the precise solutions. I did mention the denominator equals to 0 as another solution though haha. Yeah we Asian kids are taught to be fast in the exam, that is pretty common for us. You just earn a far away sub for the channel, as my teacher used to said, math is a universal language, no matter he or she comes from.🤣please keep up the nice job!
Cheers mate. But at the same time, 3^x+1 cannot be zero (you cannot divide by zero), so the imaginary solution is not an actual solution to que original problem, correct? Note: I didn't exactly use Euler's identity per si, in order to solve the problem and find the 2nd theoretical solution, but wrote -1 in polar coordinates instead, which is basically the same thing as using Euler's identity.
This can be solved very easly. Let me explain 9 to power of x - 1 can be split into. (3 to power of x +1) ( 3 to power of x -1) Finally we are getting. 3 to power of x - 1 = 8; then 3 to power of x-1= 3 to square; x =2 ;;
Si quieres aprender más de matemática física y química , sígueme.
Ok master, I am very much ready and willing to unlearn old skill and learn new ones sir. Your link please 🙏🙏🙏
i found the value of x by testing x = 2; you not only showed exact way how to find x=2, but also the second complex value of x by making clever use of Euler equation, thank you
Thanks a million for gaining values and finding our contents exciting and educative sir.
Much love from all of us @onlinemathsTV. 💕💕💕
@@onlineMathsTV Hi, i found how to get 2 of the equation: first, 9^x=3^2x, then 3^2x - 1 = (3^x+1)*(3^x-1) for a^2-b^2=(a+b)*(a-b), then 3^x+1 from up and 3^x+1 from down are gone, then 3^x-1=8, 3^x=9, x=2
Kochanieńki ! za dużo pisaniny, to równanie rozwiązałem w jednej linijce, łącznie z zespolonym, bo 9^x-1=(3^x+1)(3^x-1), a ln(-1)=Pi*i
What you did was rigorous and shed light on Euler's identity . e^πi+1=1
Yap
I dont think the second solution is right...elaborate if i miss anything.thanks
I am a Taiwanese, watching this random video at my 3AM😂While I would go to work at 9AM😂Why am I here
How about make 9^x-1 equals to (3^x+1)*(3^x-1), and then literally divide 3^x+1, we got 3^x-1=8, thus x=2; Considering 3^x+1=0 for a expectation, thus x=blablabla
Using the difference of two square for the numerator as you have rightly stated will only give the real value x=2 thereby omitting the imaginary root. In case of exams where you are asked to solve for only the real root then this approach is the fastest and the best. Thanks for your contribution and thanks for creating out time to watch our contents despite your tight schedule. We love you sir.
@@onlineMathsTV Hi, I would like to thank you for throughly replying, and also for the precise solutions. I did mention the denominator equals to 0 as another solution though haha.
Yeah we Asian kids are taught to be fast in the exam, that is pretty common for us.
You just earn a far away sub for the channel, as my teacher used to said, math is a universal language, no matter he or she comes from.🤣please keep up the nice job!
@@onlineMathsTV the imaginary root is wrong cuz 3^x cant be -1 as this would end up with 0 in denominator
Cheers mate.
But at the same time, 3^x+1 cannot be zero (you cannot divide by zero), so the imaginary solution is not an actual solution to que original problem, correct?
Note: I didn't exactly use Euler's identity per si, in order to solve the problem and find the 2nd theoretical solution, but wrote -1 in polar coordinates instead, which is basically the same thing as using Euler's identity.
You are right @Luisgoncalves and thanks for the second approach suggested.
Respect sir 🙏🙏🙏
This can be solved very easly. Let me explain
9 to power of x - 1 can be split into. (3 to power of x +1) ( 3 to power of x -1)
Finally we are getting. 3 to power of x - 1 = 8; then 3 to power of x-1= 3 to square; x =2
;;
Bravo...👍👍👍
You the best sir.
Respect.
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