A System of Exponential Equations from Cambridge A-Level Exam
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- Опубликовано: 9 авг 2022
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You sound a bit unwell Syber. Hope you get better soon.
Yeah his voice isn't as exuberant and enthusiastic as it is always, get well soon Syber ❤️❤️
Thank you! 🥰
Thank you very much! ❤️
I used Method 1. Many people will just jump to using natural (base e) or common (base 10) logs, because of familiarity. But a good rule of thumb is to choose a base that makes some terms go away and simplifies as much of the calculation as possible. Base 6 in this case.
your daily dose of maths videos are soo much helpful..thnk you sybermaths
Glad you like them!
2^x = 3^y, x+y= 1
If x+y= 1, that means x= 1-y
2^(1-y) = 3^y
=> 2/(2^y)= 3^y
=> 6^y= 2
=> y= Log_6(2)
Since x= 1-y, x= 1-Log_6(2)
So x= 1-Log_6(2), and y= Log_6(2) are the solutions
Oh wait I realised that it can be written your way
1 = log_6(6) so if we simplify x further then x = log_6(6) - log_6(2) = log_6(6/2) = log_6(3)
You are right
This time I solved it by the first method.
But the 2nd one is, as usual, more interesting.
Congratulations. 🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷
Thanks! 😃
Given:
2↑x = 3↑y - ①
x + y = 1 - ②
To find:
x, y
As it can never be zero, multiplying both sides of ① by 2↑y:
(2↑x)·(2↑y) = (3↑y)·(2↑y)
Using『(a↑b)(a↑c) = a↑(b + c)』and『(a↑b)(c↑b) = (ac)↑b』to rewrite both sides:
2↑(x + y) = (2·3)↑y = 6↑y
Substituting with ②:
2↑1 = 6↑y
6↑y = 2
Taking log (base 6)「lh」on both sides:
y = lh(2)
x + y = 1
x = 1 - y
x = 1 - lh(2)
Using『1 = lh(6)』and『log(a) - log(b) = log(a/b)』to rewrite x as a single logarithmic term:
x = lh(3).
x = lh(3) and y = lh(2) where lh is log (base 6).
(log_6(3), log_6(2))
Dear teacher! Big request:
explain in more detail. This will be very helpful for beginners and will not be torture for more experienced students. for many it is very important! Thanks in advance and good luck with your endeavors!
I agree with you. Sometimes I try to keep the videos short and tend to rush
In method 2, both x and y can be found the same way. To find y, multiply both sides by 2^y (like you did), and to find x, multiply both sides by 3^x.
Thanks for sharing
I'd like to pose a geometric challenge:
Find a trapezium (two parallel sides) so that the four sides and the longer diagonal follow a geometric progression starting with 1.
So if the four sides are a, b, c, d and the longer diagonal is D, the lengths would be a=1, b=r, c=r^2, d=r^3 and D=r^4.
Find the exact value of the ratio r and the area of the trapezium.
Let ABCD the trapezium with AB , DC parallels and DH the height of trapezium : Area = 0.5*DH*(AB+DC)
AD² = DH² + AH² and BD² = DH² + BH² etc....
Heron formula for triangles ADB, BCD giving : Area trapezium = area ADB + area BCD ...
I think we can do the exact reverse over here 3:50
2^x × 3^x = 3^y ×3^x
6^x=3^(x+y)
6^x=3
x=log(base6)3
That's really cool!
Your voice sounds a bit cracked, don't forget to take care of yourself Syber
Thank you! 🥰
Wow. I am proud to be a 10th grader to solve this. Thank you so much and please take care.
Beautiful and simple question.
hey boy take care👋🎩
Thank you! 🥰
Hey finally one I was able to solve on my own!
Awesome! 💖
Very easy 👍😜
2^x = 3^y
2^(x+y) = 2 = 1.5^y * 2^y * 2^y = 6^y. y = log (6, 2). 6 ≠ int^n, so if we want to see log(6,2) like a n/m we can't. So it's irrational that = 0,3868... So, x = 1 - log (6,2)
👍
Are you ok bro?
Sore throat 😷
method 100 : Recall a^x = e^x*ln(a) ---> 2^x = e^x*ln(2) and 3^y = e^y*ln(3) ---> x*ln(2) = y*ln(3) etc...
just ln both sides. A lot easier...
@@SyberMath I knew that just to show another way (change of exponentiation base) !
Using first part of 3rd method I got here . . . x/y = ln3/ln2 . . . subbing 1-y for x I get an equation in y . . . namely (1-y)/y = ln3/ln2 . . . now if I multiply both sides by y and factor my equation becomes . . . y(y+0.585) = 0 . . . . . . . . but this does not yield a correct solution, and I can't see my error. Can anyone help, thanks
if you solve for y, you should get y(ln(3)+ln(2))=ln(2) and
y=ln(2)/(ln(3)+ln(2))
@@SyberMath Thanks for taking the time to reply, but I still can't see my error
I used a method similar to your third.
Nice work!
thank you so much but is it just me or i think your sound unwell
i hope it is not but if it is get well soon friend!
Thank you! 🥰💖
I have sore throat
@@SyberMath oh get well soon Syber!!!
Hey SyberMath, what university did you attend?
I hate how “difficult” math problems are so easy just so it’s actually possible
😁
Syber..... Sore Throat I guess?
Take care Love you
Yes. Thank you! 🥰
what is A-level ?
"The Cambridge Advanced Level or A Level is a pre-university programme conducted by the prestigious Cambridge Assessment International Education (CAIE), United Kingdom. The Cambridge A Level programme is the gateway to the most prestigious universities worldwide."
What happened to your voice?😁
I lost it!
@@SyberMath you are always working, so energetic.
Despite the fact that you lost your voice, you still made a video for us. Thank you so much and quick recovery.