That was superb Professor. I was just about able to follow it, but I shall need to watch it a number of times to maximise the learning. Thank you so much! Really enjoyable!
if you substitute a=3x then the equation becomes 2^a=2a which has solutions at a=1 a=2 but does all the algebra amount to more than a guess. 2^a/2a =1= 2/2 or 4/4 or taking logs a-log_2(a)=1 =1-0 or 2-1
That was superb Professor. I was just about able to follow it, but I shall need to watch it a number of times to maximise the learning. Thank you so much! Really enjoyable!
You can do it! Glad it was helpful! Excellent! Merry Christmas 🎉🥳🎁
if you substitute a=3x then the equation becomes 2^a=2a
which has solutions at a=1 a=2 but does all the algebra amount to more than a guess.
2^a/2a =1= 2/2 or 4/4
or taking logs
a-log_2(a)=1 =1-0 or 2-1
You can see that for x = 0,667, it is close enough. So x = 0,667
As per Lambert W function
X=1/3....
Graphic solution !?
I don't get the Lambert function but x = 1/3, x = 2/3.
1=6x/8^x , /:2 , 1/2=3x/2^(3x) , 1/2=3x*e^(-ln2*3x) , /*-ln2 , -ln2/2=-ln2*3x*e^(-ln2*3x) , W(-ln2*e^(-ln2))=W(-ln2*3x*e^(-ln2*3x)) ,
-ln2=-ln2*3x , x=1/3 , / W(-ln2/2)=-ln2*3x , x=W(-ln2/2)/(-ln2*3x) , x=1/3 / , test , 8^(1/3)=2 , 6*1/3=2 , OK ,
8^×=6×
=>1/2=3×.e^-2^3×
=>-(2^-1)ln2=(-3×)ln2.
e^-3×
=>W{(-1)ln2.e^ln2(-1)}
=W{-3×.ln2.e^ln2(-3×)}
=>-1ln2=-3×ln2
=>×=1/3
=