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What is the value of x in this equation??
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4^x=u 16^x=u^2 64^x=u^3 And we have a simple equation
Have you tried solving it like that? Could you please explain?
As he said, then u + u² = u³ u³ - u² - u =0 u(u² - u - 1) =0 First solution of u u =0 u² -u -1 =0 ∆= 5 u = (1±✓5)/2 4^x = (1±✓5)/2 x log 4 = log ((1±✓5)/2) (log ((1-✓5)/2)) Is not valid x log 4 = log ((1+✓5)2) x = (log (1+✓5)/2)/log 4
This does not check. Try quadratic form.
It's official, Math is gonna give me PTSD in college 💀💀💀💀
Equations like this can be solved by first converting to a quadratic equation. 1^x=1, 5^x=a, 25^x=(5^x)^2=a^2. 1+a=a^2. Then a^2-a-1=0 Using quadratic equation to solve for a you get a=(1+5^.5)/2 a=5^x=(1+5^.5)/2. Then ln 5^x = ln((1+5^.5)/2). x=ln((1+5^.5)/2)/ln5=.2989937178. Check 1^.2989937178 + 5^.2989937178=25^.2989937178. 🎆🎉
Accent is funny
Excellence
Nice question my foot 🤡. Which standard is this problem even for lol
(7^x)⁰ + (7^x)¹ = (7^x)² -> 7^x = (1±√5)/2 ...
can't you substitute y=7^x in the first step to get 1+y=y^2? I mean, since we are looking for real solutions, we can say that 1^x is always 1, no matter the x
It's too much
Nice demonstration
Okay, this is how I would solve this equation. Step one, move all the terms to the left side of the equation so that you have an equation equal to zero 4^x +10^x -25^x =0. Factor out 4^x so that you have 4^x(1+2.5^x -6.25^x)=0 Let 2.5^x =a. Then 6.25^x=2.5^2x and therefore equals a^2. So then 1 + a -a^2=0. Use quadratic equation to solve for a and you get a= (-1-5^.5)/-2=(1+5^.5)/2=1.618033989 ln a=xln 2.5 =ln 1.618033989. Then x=ln 1.618033989/ln2.5=.525173734 Check 4^525173734 +10^525173734=25^525173734 Q.E.D 🎉🥳
I would always solve these equations in two steps. 1^y=1. 2^y=a and 4^y=2^2y=a^2. So now you have a quadratic equation 1+a=a^2. a^2-a-1=0. Using quadratic formula you get a=(1+5^.5)/2=1.618033989. End of step 1. Step 2 ln a = ln2^y=yln 2=ln 1.618033989 and therefore y=ln1.618033989/ln2=.6942419136. y=.6942419136. Check 1^.6942419136 +2^.6942419136=4^.6942419136 Q.E.D. 🎉🎆
Yo dude for the first question I managed to get one of x's solution to be x=root10(9), if you plug it in back into the equation you'll get back 36. Also nice vid btw 👍
3^17 = 129140163 :)
This can be solved as a quadratic equation. Let 6^x =a. Then 36^x=6^2x=a^2 and 1^x=1 You then have 1 + a=a^2. a^2-a-1=0 a=(1 +5^.5)/2=1.618033983=6^x Then x=ln1.618033983/ln6=.2685694332 Plug x into the original equation and it verifies this answer.
Would x= log ((1+√5)/2) be an acceptable 8 answer.
Oh I thought the box said x=2 and then I watched the whole video looking for the mistake. Cool problem in the end tho
you have an extremely unique handwriting! At first I thought the answer would clearly be x=2 but after a few seconds i got hooked on this video
Why cant u do this, [(1/2)^y]^2 = (1/2)^2y So, (1/2)^2y + (1/2)^y = 1 And, 1 = 1/2 + 1/2 So, (1/2)^2y + (1/2)^y = (1/2)^1 + (1/2)^1 Since all bases are similar, 2y + y = 1 + 1 3y = 2 Y = 2/3
Hat der Logarithmus keine Basis?
are they really Math Olympiad questions? Can't believe how easy they are
wrong answer.
the answer is log(1/2 (1 + sqrt(5)))/log(9)
I am terrible at Maths but I did a little bit of trial and error using my calculator and got the result as 1.25 (very roughly). What am I missing here? How can the answer be 9 (I skipped to the end to see you arrived at 9).
7
Hello man, amazing video! Thanks I need to share a fun fact about the first equation I had tried to solve this before I saw the video and find this: x = ln (phi)/ln(4) or just log(phi) in base 4 , when phi is the golden ratio After I saw the video, I lost my way when KPA found sqrt(5)-1 and not sqrt(5)+1 I lost my shit in that, trying to find what I did wrong But I forgot that this goddam golden ratio has good tricks and so does its powers I know that I just needed to find the decimal form of the both numbers (the one I found and the one that the KPA found) But i was so sure that him or me did something wrong (probably me of course, and indeed I was kind of wrong), some assumption or maybe some wrong calculation I only noticed the both were equal when I assumed their equality and I found the phi²-phi-1=0, this equation is true for the golden ration, by the fact that the golden ratio is one of its roots I sit here for almost a hour thinking about it (it's 02:35 AM in Brazil, for God sake kkkkkkkkk) , I know it's easy to notice that (just plot them on the calculator, your bastard), but it was a great and goofy way to find this Saying all that, I can tell you guys: log( ( sqrt(5)+1 )/2 ) in base 4 = log( ( sqrt(5)-1 )/2 ) in base 1/4 www.wolframalpha.com/input?i2d=true&i=is+Log%5B4%2CDivide%5BSqrt%5B5%5D%2B1%2C2%5D%5Dequals+to+Log%5BDivide%5B1%2C4%5D%2CDivide%5BSqrt%5B5%5D-1%2C2%5D%5D%3F Thank you for your time, I need to sleep now!
ln(1/2) = -ln(2). You could have made this simplification to expres the result in log base 2. Anyways, good pedagogy.
Thanks
Learn to recognize Golden Ratio problem: 2^x=Φ=(√5+1)/2, so x=lnΦ/ln2=0.69424... How? Divide the given equation by 18^x and rearrange to 2^x-1-1/2^x=0 Then multiply by 2^x: (2^x)^2-2^x-1=0 and substitute Φ=2^x to obtain Φ^2-Φ-1=0 The positive root of this equation is Φ=(√5+1)/2 as claimed.
*What about complex solutions?*
You can't divide when the sign is addition or minus
See I ain't solving that my lazy brain would just tell me to do hit and trial ie x³-x²=100 then we will take x²as common we get x²(x-1)=100 and clearly our x is 5😂
that's what we do in math methods
For (1/9)^x=(-1-(5)^(1/2))/2, x is a complex number! Why dismiss that one?
So just the real solutions?
Very nice presentation of the problem. Thanks ks for sharing
log
Dude, you need practice. x is approximately 0.525 This is one of those you can do visually just because you can see they share 100 as a lowest common denominator.
4^x=y 1+y=y² y=φ or y=1-φ, where φ=(√(5)+1)/2 x is a value such that 4^x=φ or 1-φ. If x must be real, x=ln(φ)/ln(4) Otherwise, x is either (ln(φ)+2πNi)/ln(4), where N is some integer, or (ln(1-φ)+2πNi)/ln(4). ln(1-φ)=-ln(φ)+πi. So the numerator either contains +ln(Φ) and an even multiple of πi, or -ln(φ) and an odd multiple of πi. Thus, x=((-1)^N*ln(φ)+πNi)/ln(4), where N is any integer. The implication, of course, being that there's no ∞ trickery and that 1^x is always just 1.
You really stand up to your name, "mathmachine".
I like maeth but without the e
wasn't it better to solve log(1/10) which is equal to -1 so you have as a final answer -log((√5-1)/2)?
Thank you sir
too easy, what is this.
1^x = 1; let 9^x=y, 1+y^2=y and then take log 9 y The first few steps are not necessary
Why not just 5^(2x)-5^x-1=0 5^x=y Y^2-y-1=0 (1±(5)^1/2)/2=y=5^x Log5((1+(5)^1/2)/2)=x No fractional logarithms this way, no messing with fractions at all really. Just clean and simple
A solução real é o 3. Há 2 soluções complexas:-3+iV23 tudo sobre 2 e -3-iV23 tudo sobre 2.
It is not possible to understand your voice
Sorry but your math is wrong 4/25 is 0.16 2/5 is 0.4 so your simplification is incorrect
a=0 is also a solution
No it's not because any real number other than 0 to the power of 0 is 1 so here we will have 2⁰+4⁰=8⁰===> 1+1=1 but that's obviously wrong
@@yassinechronos3829 oh yeah.