A Cool Exponential Equation (a video response to
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- Опубликовано: 21 авг 2022
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ashamed to admit that with a BS in Engineering and an MS in Applied Math, I did not quickly see how to work this out. In my defense, I am now 65
Easy on yourself man
😂😂
It happens sometimes even to mathematics teachers
No worries! You'll get there
if you follow the algebra the answer will find itself !
By inspection, we can easily see that x = 1 is one root, and by taking derivatives, we can see that there can't be any roots for x > 1. Take natural logs of both sides of the original given equation, and we get x^2 ln 3 + x ln 2 - ln 6 = 0, which is a quadratic equation in x. Because it's a quadratic equation, the sum of the roots must = - (ln 2)/(ln 3). Since we already know that x1 = 1 is a root, to find the other root x2, x1 + x2 = - (ln 2)/(ln 3), so x2 = - (ln 2)/(ln 3) - 1, which is approximately -1.6309
You don't need any quadratic equation my friend. Here is my solution, way faster:
2^(x-1).3^(x²-1)=1.
(2.3^(x+1))^(x-1)=1
This can be true if x-1=0 (x=1) or if 2.3^(x+1)=1 (1)
(1) => log(2)+(x+1).log(3)=0 => x+1=-log(2)/log(3) => x=-log(6)/log(3).
Problem destroyed.
@@italixgaming915 isn't supposed to be 2.3^(x-1)=1
Since x=1?
Given:
(2↑x)·(3↑x²) = 6
To find:
x
Rewriting 6 as 2·3:
(2↑x)·(3↑x²) = 2·3
As 2↑x and 3↑x can never be zero, dividing through to separate primes:
(2↑x)/2 = 3/(3↑x²)
Rewriting fractions as sums of exponents:
2↑(x - 1) = 3↑(1 - x²)
Taking natural log on both sides:
(x - 1)·ln(2) = (1 - x²)·ln3
moving all terms to LHS:
(x - 1)·ln(2) - (1 - x²)·ln3 = 0
Rewriting (1 - x²) as -(x² - 1):
(x - 1)·ln(2) + (x² - 1)·ln3 = 0
Factoring x² - 1:
(x - 1)·ln(2) + (x - 1)·(x + 1)·ln3 = 0
(x - 1)·(ln(2) + (x + 1)·ln(3)) = 0
from here, x = 1, or ln(2) + (x + 1)·ln(3)
Distributing x + 1 and rearranging:
x·ln(3) + (ln(3) + ln(2)) = 0
Rewriting ln(a) + ln(b) as ln(ab):
x·ln(3) + ln(6) = 0
x·ln(3) = -ln(6)
x = -(ln(6)/ln(3))
x = -lt(6), where lt is log with base 3.
Aww it is soo haarddd
@@mathlove7474 Don't whine, this is high-school level mathematics. It's not easy either, but once you know how exponents and logarithms work you should find this straightforward. Practice and a willingness to learn are all you need to understand this level of complexity.
It's also why I try to explain every step I take in these. So kids still learning these ideas can follow along.
Wow, I answered 1 immediately, without taking in consideration other solutions! I have to open my mind and be more elastic, this is the lesson I received, valuable in all aspects of life.
Excellent!
Wow, 2nd method makes you to go far more in depth than 1st one. Great!
You guys are doing great job @blackpenredpen and @SyberMath
Thank you!
I really hope you are doing fine. I am so impressed at the fact that you do keep making videos, even if your health does not upto the mark. Kudos my friend.
I'm doing fine! Thank you for the kind words!!! 💖
You should have put parenthesis for x^2 to indicate that it’s not 3^2x.
Very nice procedures. Thank you!
Glad you enjoyed it!
Lavorando sull'equazione risulta log(2)3*x^2+x-log(2)6=0,che é una semplice equazione di 2 grado che dà 2 soluzioni, x=1,x=-1,631
I havent been taught logarithms yet, but this was explained well enough that i still understood it!
Glad to hear that!
@@SyberMathjust don’t assume people know what these things mean wtf is a log
I found out in the first glance that logarithm is going to work.
The given equation is rewritten as 2^x(3^x)(3^(x^2-x))=6, then we have (6^(x-1)(3^(x^2-x)=1, both sides take log, we have x = 1 or x = -ln6/ln3
This question requires a bit of thinking but it's not so bad once you know how.
3rd method?
2^(x-1)=3^(1-x^2)
1st solution when the exponent s (x-1) and (1-x^2) are contemporary equal to zero. This is right if x=1.
The 2nd solution when x is different by 1. So we can simplify elevating all members to 1/(x-1).
We have 2=3^[-(1+x)]
2=(1/3)^(x+1)
x+1=log in base 2 of 1/3
x=(log in base 2 of 1/3) -1
Changing base to the log
we have x=-1-log in base 3 of 2
Is it (3^x)^2 or 3^(x^2)? They are not the same thing, I think.
At first it seemed at bit too easy: x=1 obviously works.. so that's it. But in fact you show it to be really interesting quadratic equation, and analyse it beautifully. Compliments and Thank you!
Np. Thank you! 😍
So the result is x1 =1 and x2 = -1.63091.
It's a thumbs up from me.
It's very clear, thank's very much sire.
Np.
You can see my math videos too, and learn new techniques 😃
Are you iranian?
Im iranian too, you can see my math videos, i learn there very usful techniques 😃
This is indeed a nice exponential equation. For me, the coolest part was checking the 2nd solution, x=-log₃6, in the original equation, 2ˣ×3^(x²)=6:
3^(x²)=3^(-log₃6)²=3^(log₃6)²=(3^(log₃6))^log₃6=6^(log₃6).
So 2ˣ×3^(x²)=2^(-log₃6)×6^(log₃6)=6^(log₃6)/2^(log₃6)=(6/2)^(log₃6)=3^(log₃6)=6.
Thank you!
I'm I the only one that notices the limit of subscribers is approaching 100k ?
Congratulations in advance 🎉
Thank you, sir!
Congrats on your channel, btw. You gained a quite a few subs 🎉
@@SyberMath Thank you 🙂
You're welcome! I'm glad your channel is doing well! 💖
damn im stuck after doing the quadratic formula
So, u have to know, the properties(change of base, etc) & also algebra.
I really love your videos.
Glad you like them!
Ans =1 (2to the Power 1 =2 & 3to the Power 1=3) and 2×3=6.proved.
Great.
you my favourite math guy on yt
🥰 Thank you!!! 🥰
At a glance x=1
By inspection x = 1
I did it by the second method
question for all: prove that there is no other roots
Böyle devam reis
Saolasin! 🥰
Can you do it like this
2^x . 3^x²=6
2^x .3^x²=2¹ . 3¹
Bases are the same so equate exponent to exponent
x . x²=1 . 1
x³=1
x=1
No you can't say that, but here is a valid solution:
2^(x-1).3^(x²-1)=1.
(2.3^(x+1))^(x-1)=1
This can be true if x-1=0 (x=1) or if 2.3^(x+1)=1 (1) - you forget the second case in your solution.
(1) => log(2)+(x+1).log(3)=0 => x+1=-log(2)/log(3) => x=-log(6)/log(3).
Problem destroyed.
very nice
Thanks, Nico! 💕
2^x. 3^x^2 = 6
= 2. 3
x^1 = 1
x^2 =1
answer x =1 and x=-1
they have to be the same x values
Because 2*3=6, you can see the solution with one eye without any calculation. 🤫
How will u find the other root 🤦♂️
@@adithyan9263 Hercule Poirot will find the second root !
@@WahranRai what is that🤔
Can you please try this one my friend?
integrate (x^dx -1)
yes, dx is in the power of x
doesn't it make no sense?
U took that ibe from Bri the math guy 😂
@@pranjalsrivastava3343 kinda but u can use expansion and make the dx come down
@@adithyan9263 ik this is from bri😁😁
Thanks sir 😙😌😙
Most welcome
A Cool Exponential Equation: (2^x)[3^(x^2)] = 6; x = ?
(2^x)[3^(x^2)] = 6, [3^(x^2)]/3 = 2/(2^x), 3^(x^2 - 1) = 2^(1 - x)
Convert the exponential base number 2 into 3 using logarithmic math:
Let 3^n = 2, n = log2/log3 = 0.631; 2 = 3^0.631
3^(x^2 - 1) = 2^(1 - x) = 2^(0.631 - 0.631x); x^2 - 1 = 0.631 - 0.631x
x^2 + 0.631x - 1.631 = 0; x = - 0.631/2 ± sqrt[0.631^2 + 4(1.631)]/2
x = - 0.315 ± (sqrt6.922)/2 = - 0.315 ± 2.631/2 = - 0.315 ± 1.316
x = 1.001 or x = - 1.631
Answer check:
x = 1.001, (2^x)[3^(x^2)] = (2^1)[3^(1^2)] = 6; Confirmed
x = - 1.631, [2^(- 1.631)]{3^[(- 1.631)^2}
= (0.323)(3^2.660) = (0.323)(18.584) = 6; Confirmed
Final answer:
x = 1 or x = - 1.631
the original questuion was (2^x)(3^x^2) = 6 not (2^x)[3^(x^2)] = 6
I don’t understand the first part, because I don’t speak English like a normal way
Google translate hlelp you
@@mathlove7474 Que mala
Well This is not the usual substitution on steroid we see here but still interesting
If X×3^(log[2](x))²=6
Find x ?
Im iranian tooo, morteza😂😂
@@mathlove7474 🇮🇷 Iran's flag is high
X=1
X=6/6=1
I like it 😊
Good to hear
This true for 3^×^2
Why you didn't write 6 as 6.1 you'll find other solutions
We go off of prime factors
👍
X =1
Very easy ez
x=6
x = 1
X=1 yaaa. Çok kasmaya gerek yok...
Aww, come on!!! 😂
@@SyberMath 🤣🤣🤣🤣
🤓
like
用观察法得知x=1
Х=1