You are extracting an exception when there are two numbers involved. However when there is one variable and one number , then both a and b have to be the same. 2 and 4 are not the same but if x^2=2^x, then x has to be equal to 2.
@@JacobHa yep. But that's just one aberration among infinite other solutions. The accuracy will still be (100-1/♾️)%.For all real purposes, it means 100%.
To show that 1/2 is the only solution you can use derivatives since √x is always increasing in it's domain and (1/2)^x is always decreasing :)!!!
x = 1/2 . . . By inspection, obviously. But is it the sole answer?
For real numbers, yes. But if you're including complex numbers, then all branches of W(ln(4))/ln(4) are also solutions.
I did: x^(1/2x) = 1/2, hence x=1/2.
2:00 Here, unless my ears are playing tricks on me, you said “0 IS a solution”, after explaining how it cannot be.
how do you prove an increasing & decreasing function have most 1 soln?
I also got x=1/2.
Х=1/2 виден без решения.
x = 1/exp(W(ln(4))) = 0.5
5:10 Do we need Lambert's W function?
(2x ln2) e^(2x ln2) = (ln2) (2)
(2x ln2) e^(2x ln2) = (ln2) e^(ln2)
Therefore, 2x ln2 = ln2, which gives x = 1/2
a^b = b^a does not imply a=b
for example take a=2 and b=4
You are extracting an exception when there are two numbers involved. However when there is one variable and one number , then both a and b have to be the same. 2 and 4 are not the same but if x^2=2^x, then x has to be equal to 2.
@reconquistahinduism346 x=4 is also a solution
@@JacobHa yep. But that's just one aberration among infinite other solutions. The accuracy will still be (100-1/♾️)%.For all real purposes, it means 100%.
x = 1/2
x = 1/2
(1/2)lnx=-ln2 x...lnx=-xln4...lnx=t..t=-ln4e^t..(1/t)(-ln4)=e^(-t)..(-t)=W(ln4)...x=e^(-W(ln4))
1/2 * lnx=x*ln(1/2) , ln(x)/x=2*ln(1/2) , lnx*e(-lnx)= -2*ln(2) , /*(-1) , -lnx*e(-lnx)=2*ln(2) ,
-lnx*e(-lnx)=ln(2)*e^(ln2)) , -lnx=ln2 , ln(1/x)=ln2 , 1/x=2 , x=1/2 , test V=(1/2)=V2/2 , (1/2)^(1/2)=V2/2 ,
x = 1/2