Question

Find all the real-number roots of the equation. Give an exact expression for the root.

e^x+e^-x=2

Answer 1

\[e^x-2+e^{-x}=0\]

multiply through by \[e^x\]
\[e^{2x}-2e^x+1=0\]

let \[u=e^x\]

then we have

\[u^2-2u-1=0\]

solve for u...return back to x

Answer 2

mistake:
should be
\[u^2-2u+1=0\]

Answer 3

why can't i just subtract [e ^{x}\]

Answer 4

?

Answer 5

e^x

Answer 6

i have an idea. solve
\[z+\frac{1}{z}=2\] and the replace \[z=e^x\]

Answer 7

\[z+\frac{1}{z}=2\]
\[z^2+1=2z\]
\[z^2-2z+1=0\]
\[(z-1)^2=0\]
\[z=1\]
\[e^x=1\]
\[x=0\]