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An Exponential Equation That Will Make You More Brilliant
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- čas přidán 2. 07. 2024
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How did you get that black board app where you can write on it.
@@AlanGarfield-rz9wu Notability
@@SyberMath thank you ♥
Let's go!!!! Finally a sponsorship! You have grown so much man! What a journey.
Thank you 🥰
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@@SyberMathtry this problem
Floor function (2x-3)+absolute value(x-2)=x-3
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
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.
Thanks!
(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
My 🐐 has finally got a sponsor 🙌
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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
Nice problem - and congrats on the sponsorship!
Thanks! 😍😊
Brilliant expands its kingdom
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
I think you can find the third solution by this way. x^2=2^x=e^(i*2pi*n)*2^x...
Have you verified x=e^(-W(-(πi(1/2+n)+ln(2)/2))) by calculation? You can do that with Wolframalpha.
Congrats on the sponsor, man!
Thank you!
x = 2
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
Any programmer will likely see this quickly.
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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 < -ln(2)/2 < 0 ===> 2 real solutions
x = e^(-W_[0](-ln(2)/2)) = 2 #
x = e^(-W_[-1](-ln(2)/2)) = 4 ##
2nd case: x < 0
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)
Wow!
Really cool.
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