Bonita questão, hein professor? Gostei do seu desenvolvimento. Ainda que eu não entenda bem o Inglês, porém, dá para acompanhar a sua explicação. OK? Deus abençoe você professor.
The domain of the given expression clearly indicates that 3^x < 5. You cannot have ( 5 + √21)/2 as a value of 3^x. Am I missing out on something here ? Kindly review.
Alternative sol: Note: log will signify a base 3 logarithm for the rest of the problem. Let x=logy log(5-3^logy)+logy=0 log(5-y)+log(y)=0 log(5y-y²)=0 5y-y²=1 y²-5y+1=0 Σх=Σlogy=logΠy. The product of y1 and y2 by vieta is 1, therefore the sum of roots for x is log1=0
I loved what you did here. Thanks for the great video.
What a nice problem with a nice, unique and detailed/explicit explanation from Onlinemathstv.
More wins sir....❤🎉
Nice problem. Was a little bit of work. Well explained, thanks for the lesson and video. 👍
log₃(5 - 3ˣ) + log₃3ˣ = 0
3ˣ(5 - 3ˣ) = 1
3ˣ = u => x = log₃u
u(5 - u) = 1
u² - 5u + 1 = 0
u = (5 ± √21)/2
*x = log₃[(5 ± √21)/2]*
nice! I was also thinking about this
❤
Excellent video, greetings from Monterrey, Mexico.
Bonita questão, hein professor? Gostei do seu desenvolvimento. Ainda que eu não entenda bem o Inglês, porém, dá para acompanhar a sua explicação. OK?
Deus abençoe você professor.
t=3^x, 0
The domain of the given expression clearly indicates that 3^x < 5. You cannot have ( 5 + √21)/2 as a value of 3^x. Am I missing out on something here ? Kindly review.
(5 + √21)/2 < (5 + √25)/2
(5 + √21)/2 < (5 + 5)/2
(5 + √21)/2 < 5
@@zeroone7500 Thank you so much. I had missed out on this.
Nice chalk :), had fun with this
Hahahahaha.....😍😂😂🤣😍🤣🤣
Thanks sir
Yes nice problem.
Alternative sol:
Note: log will signify a base 3 logarithm for the rest of the problem.
Let x=logy
log(5-3^logy)+logy=0
log(5-y)+log(y)=0
log(5y-y²)=0
5y-y²=1
y²-5y+1=0
Σх=Σlogy=logΠy.
The product of y1 and y2 by vieta is 1, therefore the sum of roots for x is log1=0
Wonderful approach from you sir.
/// if 3^x>5 , x=(ln(5+V21)/2))/ln3 or x=log₃((5+V21)/2) not a solu ///,
solu , x=((ln(5-V21)/2)/ln3 , or x=log₃((5-V21)/2) ,