Question

0=e^(2y)-e^(y)

Need to solve for y here, don't understand why answer is -ln(2) in the book. Thanks.

Answer 1

woops there is a 2 in front of the first e&(2y).

Proper equation: 0=2e^(2y)-e^(y)

Answer 2

treat it like a quadratic equation ...

let e^y = x
then your equation becomes
0 = 2x^2 - x
x(2x - 1) = 0
x = 0 or x = 1/2

then
e^y = 0 -----> no solution
or
e^y = 1/2
take ln of both sides and use the rules of logs
y lne = ln1 - ln2
y (1) = 0 - ln 2
y = -ln2

or

e^y( 2e^y - 1) = 0
e^y = 0 --> no solution as before
or
e^2 = 1/2 ----> proceed as above