I'll Be Proud If You Solve This
ฝัง
- เผยแพร่เมื่อ 14 พ.ค. 2023
- 🎓Become a Math Master With My Intro To Proofs Course!
www.udemy.com/course/prove-it...
►SUPPORT ME BY BECOMING A CHANNEL MEMBER
/ @brithemathguy
#math #brithemathguy #algebratricks
This video was partially created using Manim. To learn more about animating with Manim, check out:manim.community
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.
![[TH] VCT Masters Shanghai Playoffs DAY 6 // G2 vs GEN | 100T vs TH](http://i.ytimg.com/vi/M6jxF80AajI/mqdefault.jpg)
Check out and consider donating to Connect Me!
connectmego.org/
I find it slightly more intuitive to let a = 5^x and b = 2^x, then you'll get the same factoring situation.
Bruh🗿
Amazing 🎉
I got it! This is pretty much the only one I've ever gotten from your channel this quickly.
This is one of the basic exercises for grade 12 in Vietnam
great
Like everywhere, typical high school stuff
10th grader
Basically Asia
Lmao an average 10th grader in India can solve this easily.
damn i hope i could someday solve this types of problems in no time.
I tried to divide them up into prime-factors at first, but I did in the end get x = log_5/2 (2)
really nice, using nearly all of the power tricks
i went with logs in earlier steps and got wrong answer or made some mistake somewhere
I think you could improve the final answer a bit more by using logarithms with whole number bases
I legit didn't expect it to be easy
You always present us with mind boggling equations!
Divide by 4^x solve quadratic x= ln2/ln5-ln2 hope thats right anyway that was a nice question 😉
Awesome video, Brian! What do you think about sharing the manim code of your videos in exchange for a donation on patreon or somewhere else?
16^x - 12^x = 9^(x+y)
3^y = √2
x=?
Be proud of me, I got this one. Slightly different approach: a = 5^x and b = 2^x. Moving everything to one side gives a^2 + ab -6b^2 = 0, which factors to (a - 2b)(a + 3b) = 0. Since 5^x > 0 and 2^x > 0 we know a + 3b = 0 has no real solution. Therefore a - 2b = 0 gives the solution. 5^x - 2*2^x = 0 leads to the same answer, or since I have weird ln reflexes mine says the equivalent ln 2 / ln (5/2).
Yall saying its challanging meanwhile its the most casual excersise for 11th grade in Armania 😂
I divided both sides by 25^x and let t = (2/5)^x (t>0) and then solved the quadratic equation ( 1 + t = 6t^2 ) then I found t = 1/2 and -1/3 ( which is eliminated because t>0 ), if t = (2/5)^x then (2/5)^x = 1/2 => x = log_2/5 (1/2), that's my way!
In India this is very basic for a grade 11th students preparing for jee (joint entrance exam )
Yes
They wouldn’t even ask it in mains tho
25^x + 10^x = 6 x 4^x => (5/2)^(2x) + (5/2)^x - 6 = 0 => (5/2)^x = (-1 +/- √(1+24))/2 = (-1 +/- 5)/2 = 2 or -3. Taking logs on both sides, x = ln(2) / (ln(5) - ln(2))
I solved it using natural logs (if that's valid).
log _base 5/2_ ??
I mean yes but no one uses that. The natural log is a better choice here. x = ln(2)/(ln(5) - ln(2))
They are the same thing due to the change of base formula. log_5/2 is more direct while ln would be more useful if we actually wanted to calculate an approximation.
@@Ninja20704 I know they're the same thing, my point was that while technically correct, no one uses bases that way. But yeah maybe it's just a habit from the need to often calculate an approximation. Still feel like it's incomplete writing base 5/2 x)
The answer to this question is 0.4307 to 4 decimal places, as a result of log2/log5=0.4307 that is the correct answer
No it’s not
@@darthmaul197 yes it is
Here in the first minute of upload
Easy
This equation has only one real solution, but it has infinitely many
complex number solutions. Here is how to calculate all solutions, and
also how to calculate a natural logarithm of a negative real number.
(25^x + 10^x)/3 = 2^(2x+1)
(5^x * 5^x) + (2^x * 5^x) = 6(2^x * 2^x)
5^x * 5^x 2^x * 5^x
-------------- + ------------- = 6
2^x * 2^x 2^x * 2^x
((5/2)^x)^2 + (5/2)^x - 6 = 0
t = (5/2)^x
t^2 + t - 6 = 0
t = (-1 +- 5)/2
(t + 3)(t - 2) = 0
Real solutions:
(5/2)^x = -3 (no real solutions)
(5/2)^x = 2
x = ln(2) / ln(5/2) = ln(2) / (ln(5) - ln(2))
Complex solutions:
e^(i*pi) = -1 (Euler's identity)
e^0 = 1
C = i*2*pi*n where n E Z
e^(i*pi + C) = -1
e^(0 + C) = 1
(5/2)^x = -3 = 3 * e^(i*pi + C)
x = ln(3 * e^(i*pi + C)) / ln(5/2)
x = (ln(3) + i*pi + C) / ln(5/2)
(5/2)^x = 2 = 2 * e^(C)
x = ln(2 * e^(C)) / ln(5/2)
x = (ln(2) + C) / ln(5/2)
Answer:
x = (ln(3) + i*pi + i*2*pi*n) / ln(5/2) where n E Z
x = (ln(2) + i*2*pi*n) / ln(5/2)
if n = 0:
x = ln(2) / ln(5/2) (the only real solution)
I got the answer through a different method (completing the square). This one was fun!
Here’s my LaTeX in case anybody wants to look, though it’s a bit messy in a comment:
25^x + 10^x &= 3 \cdot 2^{2x + 1} \\\\
5^{2x} + 2^x 5^x - 3 \cdot 2^{2x + 1} &= 0 \\
(5^x)^2 + 2 \cdot 2^{x-1} \cdot 5^x - 24 \cdot (2^{x-1})^2 &= 0 \\
(5^x + 2^{x-1})^2 &= 5^2 (2^{x-1})^2 \\
5^x + 2^{x-1} &= 5 \cdot 2^{x-1} \\
5^x &= 4 \cdot 2^{x-1} \\
5^x &= 2 \cdot 2^x \\
x &= \log_5 (2) + x \log_5(2) \\\\
\therefore x &= \frac{\log_5(2)}{1-\log_5(2)}
i tried it where did i go wrong
((25^x)+(10^x))=3*(2^(2x+1))
((25^x)+(10^x))/(2^(2x+1))=3 (2^(2x+1))=((2^2x)*2)
subsitute to get
(((25^x)/2)+((10^x)/2)))/(2^(2x)=3
take 2xth root of it to get
(((25^x)/2)+((10^x)/2)))to the 2xth root/2= 3to the 2xth root
*2on both
^2x on both
use (a/2)+(b/2)=((a+b)/2)
(((25^x)+(10^x))/2)=(2^2x)*3
/3
*2
(((25^x)+(10^x))/3)=(2^(2x+1))
((25^x)+(10^x))=3*(2^(2x+1))
dang it