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

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

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

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

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

To solve the equation 0 = e^(2y) - e^y, we can use the property of exponents that states e^(a-b) = e^a / e^b.

Rearranging the equation, we have e^(2y) = e^y.

Now, let's divide both sides of the equation by e^y:

e^(2y) / e^y = e^y / e^y

Using the property mentioned earlier, we simplify the left side to e^(2y-y) = e^y, which further simplifies to e^y = e^y.

We know that e^y is never equal to zero. Therefore, to satisfy the equation, both sides must be equal to zero. This means that e^y = 0.

However, there is no value of y that satisfies e^y = 0. Therefore, there are no solutions to the equation 0 = e^(2y) - e^y.

If the book states that the answer is -ln(2), there may be an error. Please double-check the question and the answer given in the book.