e^(xcoshx)=3
xcoshx=ln3
x(e^x+e^-x)=ln9
Let: X=e^x
lnX(X+1/X)=ln9
(XlnX+1)/X=ln9
XlnX+Xln(e^(1/X))-Xln9=0
Xln((Xe^(1/X))/9)=0
So:
X=0
or: ln((Xe^(1/X))/9)=0
cancel(e^x=0), e^x is never equal to 0.
e^x*e^(e^-x)/9=1
e^(x+e^x)=ln9
x+e^x=ln(ln(9))
We cannot find the exact root, but because that x+e^x-ln(ln(9))=0 is a continuous and monotonic function and between [-1;1] this function is negative and then positive, using the intermediate values theorem, we know that there's an unique α in [-1;1] where e^(αcoshα)=0
\0/ here's our answer!
