Divide by x^ln 25 to get 1 + x^(ln 36/25) = x^(ln 30/25) ie (x^(ln 6/5))² - (x^(ln 6/5)) + 1 = 0, which is quadratic in x^(ln 6/5), and then solve this in the time-honoured way.
Great problem. Made me think for a bit. I got to the equation at 9:02 a bit faster by first rearranging as x^(2 ln 5) + x^(2 ln 6) = x^(ln 6) * x^(ln 5) and then, also dividing through by x^(2 ln 5), rearranging as x^( ln 6 / ln 5)^2 - x^(ln 6 / ln 5) + 1 = 0.
Would the other complex solution be the conjugate of the one you showed? I’m not sure that’s guaranteed, sine the original equation did not have positive integer exponents
This is way overthinking it. As observed, x=0 is a trivial solution. For 0 x^3.3 > x^ln30. For x=1: x^ln25+x^ln36 = 2 > 1 = x^ln30 For 1 x^ln36 > x^3.5 > x^ln30. Thus x=0 is the only real solution.
Divide by x^ln 25 to get 1 + x^(ln 36/25) = x^(ln 30/25) ie (x^(ln 6/5))² - (x^(ln 6/5)) + 1 = 0, which is quadratic in x^(ln 6/5), and then solve this in the time-honoured way.
I did the same, strangely though when I check the solutions I don't get same thing on both sides. Any suggestions?
@@philipfoy7117 Not sure I follow. We have
LHS = x^ln 25 + x^ln 36 = x^ln 25 . ( 1 + x^ln 36 / x^ln 25) = x^ln 25 . ( 1 + x^(ln 36 - ln 25) )
= x^ln 25 . (1 + x^ln(36/25)) = x^ln 25. (1 + x^(ln (6/5)²)) = x^ln 25 . (1 + x^(2 ln (6/5)))
= x^ln 25 . (1 + (x^(ln 6/5))² )
and
RHS = x^ln 30 = x^ln 25. (x^ln 30/x^ln 25) = x^ln 25 . (x^(ln 30 - ln 25)) = x^ln 25 . x^(ln 30/25)
= x^ln 25. (x^(ln 6/5))
and these are the same thing iff x^ln(6/5) is a root of the quadratic 1+y² = y.
Putting x^ln5=a,x^ln6=b,
a²+b²=ab
Dividing both sides by ab,
a/b+b/a=1,
a/b=m,
m²-m+1=0,
m=1±√3i/2
x=cos(π/3ln(5/6))±sin(π/3ln(5/6))
Great problem. Made me think for a bit.
I got to the equation at 9:02 a bit faster by first rearranging as
x^(2 ln 5) + x^(2 ln 6) = x^(ln 6) * x^(ln 5)
and then, also dividing through by x^(2 ln 5), rearranging as
x^( ln 6 / ln 5)^2 - x^(ln 6 / ln 5) + 1 = 0.
Bravo! Substitution transform back substitution complex solutions included.
almost made it !!! , i am getting better . my goal is being as good as you are in math , i've being following you since 2022 , thanks man
Np! You will be better. Keep up the good work 😍
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Would the other complex solution be the conjugate of the one you showed? I’m not sure that’s guaranteed, sine the original equation did not have positive integer exponents
tldr I arrived at the same solution but with a different method
x = 0
Nice - I solved for x directly though.
This is way overthinking it. As observed, x=0 is a trivial solution.
For 0 x^3.3 > x^ln30.
For x=1: x^ln25+x^ln36 = 2 > 1 = x^ln30
For 1 x^ln36 > x^3.5 > x^ln30.
Thus x=0 is the only real solution.
Here is what I did, but didn't simplify Ln(1 +/- i√3)
let x = e^y
(e^y)^Ln(25) + (e^y)^Ln(36) = (e^y)^Ln(30)
(e^Ln(25))^y + (e^Ln(36))^y = (e^Ln(30))^y
25^y + 36^y = 30^y
(5/6)^y + (6/5)^y = 1 //Divide by 30^y
let (5/6)^y = z
z + 1/z = 1
z^2 - z + 1 =0 //Multiply by z
z = (1 +/- √(1-4(1)(1))) /2
z = (1 +/- i√3) /2
(5/6)^y = (1 +/- i√3) /2
y Ln(5/6) = Ln( (1 +/- i√3) /2 ) //Ln() both sides
y = (Ln(1 +/- i√3) - Ln(2)) /(Ln(5)-Ln(6))
x = e^y = e^(Ln((1 +/- i√3))-Ln(2)) /(Ln(5) - Ln(6))
X=0, X=e^[(2κπ+-π/3)*i/(ln6-ln5)],k is an integer.