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@AlanGarfield-rz9wu

9 күн бұрын

How did you get that black board app where you can write on it.

Let's go!!!! Finally a sponsorship! You have grown so much man! What a journey.

@SyberMath

9 күн бұрын

Thank you 🥰

Ooh a sponsored video. Only the important KZreadrs can do that. :D

@SyberMath

9 күн бұрын

☺️

Congrats on the sponsorship!!

@SyberMath

9 күн бұрын

Thank you! 🥰

@user-lu6yg3vk9z

8 күн бұрын

@@SyberMathtry this problem Floor function (2x-3)+absolute value(x-2)=x-3

My 🐐 has finally got a sponsor 🙌

@SyberMath

8 күн бұрын

hehe, thanks! 😁

Nice problem - and congrats on the sponsorship!

@SyberMath

8 күн бұрын

Thanks! 😍😊

Brilliant expands its kingdom

You said that the second method was more algebraic and perhaps more rigorous. I think the first method is just as rigorous. It's easy to see that the right side is a power of 2, so by the fundamental theorem of arithmetic, x must also be a power of 2. Of course, that only works if we're looking for integer values.

@SyberMath

6 күн бұрын

Thanks!

Congrats on the sponsor, man!

@SyberMath

8 күн бұрын

Thank you!

x⁸=16^x --> x⁸=(2⁴)^x x=[(2⁴)^x]^(⅛) =2^(½x) --> x²=2^x Take natural log 2ln(x)=xln(2) --> [ln(x)]/x=[ln(2)]/2 We then may apply W function However it is simpler to note that [ln(x)]/x=[ln(2)]/2 means that x=2 But [ln(4)]/4=[2ln(2)]/4 =[ln(2)]/2 =[ln(x)]/x --> x=4 Therefore x={2,4}

x=2 x^8 = 16^x x^8 = 4^2x x^8/2x = 4 raised both sides to 1/2x x^4/x = 4 x^4/x = 2^2 x^4/x = 2^4/2 since 4/2 =2 x = 2 is a solution

(x^8)^(1/x) = (16^x)^(1/x) x^(8/x) = 16 It's easy to see that x=2 here. As a check: 2^(8/2) = 2^4 = 16

Really cool.

x⁸ = 16^x = (2^x)^4 (x²)⁴ = (2^x)^4 x² = 2^x => *x = 2* for another possible solution(s) 2lnx = xln2 (1/x)lnx = (1/2)ln2 = (1/4)ln4 so *x = 4* is also a solution

x = 2

This is how to solve this problem I will write it first x^8=16^x We move the unknowns to one side x^8-16^x=0 We get a factor like the conjugate union (x^4)²-(4^x)²=0 (X^4-4^x)(x^4+4^x)=0 x^4=4^x or x^4=-4^x We will factor it again x^4-4^x=(x^2-2^x)(x^2+2^x) x^2=2^x or x^2=-2^x Well, the answers that are clear from the symmetry of the two sides of equality can be easily extracted, such as the number two and four, but some answers cannot be determined simply. x^2=-1*2^x=e^(πi+2nπi)*2^(x) x^(1/x)=e^(πi/2+nπi+ln(2)/2) ln(x)/x=(πi(1/2+n)+ln(2)/2) ln(1/x)e^(ln(1/x))=-(πi(1/2+n)+ln(2)/2) x=e^(-W(-(πi(1/2+n)+ln(2)/2))) These are the complex answers of the equation Of course, we could find that the equation has an infinite solution x^8=16^x=e^(ln(16)x) x^(8)=1+(ln(16)x)+(ln(16)x)²)/2+...... infinity sentence a*x^(∞)+......(ln⁸(16)-1)/8!*x^8+....1=0 which must have infinite answers

@luunguyen193

8 күн бұрын

I think you can find the third solution by this way. x^2=2^x=e^(i*2pi*n)*2^x...

@82rah

8 күн бұрын

Have you verified x=e^(-W(-(πi(1/2+n)+ln(2)/2))) by calculation? You can do that with Wolframalpha.

Any programmer will likely see this quickly.

Here is how to get the results that Wolfram alpha generates: x^8 = 16^x = 16^x * exp( 2 i pi n) => x = 16^( x/8 ) * exp( i pi n/4) => x * exp( -x ln(16)/8 ) = exp( i pi n /4 ) => -ln(2)/2 * x exp( -x ln(2)/2 ) = - ln(2)/2 exp( i pi n /4 ) => - ln(2)/2 * x = W(( -ln(2)/2) * exp(i pi n /4) ) x = - ( 2 / ln(2)) * W((- ln(2)/2)* exp(i pi n /4)) n = 0,1...,7

🎉🎉🎉 sponsorship 🎉🎉🎉

@SyberMath

8 күн бұрын

😍😊

So did you prove that there are no other solutions?

x^8 = 16^x x^8 = 2^(4*x) ln(x^8) = ln(2^(4*x)) 8*ln|x| = 4*x*ln(2) 2*ln|x| = x*ln(2) ===> two cases 1st case: x > 0 x*ln(2) = 2*ln(x) ln(x)*x^(-1) = ln(2)/2 ln(x)*e^ln(x^(-1)) = ln(2)/2 ln(x)*e^(-ln(x)) = ln(2)/2 -ln(x)*e^(-ln(x)) = -ln(2)/2 W(-ln(x)*e^(-ln(x))) = W(-ln(2)/2) -ln(x) = W(-ln(2)/2) ln(x) = -W(-ln(2)/2) x = e^(-W(-ln(2)/2)) ===> -1/e 2 real solutions x = e^(-W_[0](-ln(2)/2)) = 2 # x = e^(-W_[-1](-ln(2)/2)) = 4 ## 2nd case: x x*ln(2) = 2*ln(-x) ln(-x)*x^(-1) = ln(2)/2 -ln(-x)*x^(-1) = -ln(2)/2 ln(-x)*(-x)^(-1) = -ln(2)/2 ln(-x)*e^ln((-x)^(-1)) = -ln(2)/2 ln(-x)*e^(-ln(-x)) = -ln(2)/2 -ln(-x)*e^(-ln(-x)) = ln(2)/2 W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/2) -ln(-x) = W(ln(2)/2) ln(-x) = -W(ln(2)/2) -x = e^(-W(ln(2)/2)) x = -e^(-W(ln(2)/2)) ===> ln(2)/2 > 0 ===> 1 real solution x = -e^(-W_[0](ln(2)/2)) = -0.766664695962123093111204422510314848006675346669832058460884376... # e^(-W(-ln(2)/2)) = e^(-W(-ln(2)*(2^(-1))) = e^(-W(-ln(2)*e^ln(2^(-1))) = e^(-W(-ln(2)*e^(-ln(2))) = = e^(-(-ln(2))) = e^ln(2) = 2 ## e^(-W(-ln(2)/2)) = e^(-W(-(2*ln(2))/(2*2))) = e^(-W(-ln(2^2)/4)) = e^(-W(-ln(4)*(4^(-1))) = = e^(-W(-ln(4)*e^ln(4^(-1))) = e^(-W(-ln(4)*e^(-ln(4))) = e^(-(-ln(4))) = e^ln(4) = 42^(4*x)2^(4*x)

@SyberMath

8 күн бұрын

Wow!

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