Log question?

Question

I encounter this in a hyperbolic function question.
Why e^-ln2 does not equal -2 but 1/2?

Answer 1

-ln(2)
= -1*ln2

Move the number in front of ln into the exponent of 2.
= ln(2^-1)

Simplify: 2^-1 = 1/2
= ln(1/2)

So e^-ln2 = e^ln(1/2) = 1/2

Answer 2

There are two ways to see this.

Probably the fastest is to recall that a^(-b) = 1/(a^b). So e^(-ln2) = 1/(e^ln2) = 1/2.

The other way is to recall that a*ln(b) = ln(a^b). [One of those log properties]. So -ln2 = (-1)*ln2 = ln(2^(-1)) = ln(1/2). Thus e^(-ln2) reduces to e^(ln(1/2)) = 1/2.

Answer 3

natural log rule says that -ln 2 = ln 2^-1, which becames ln .5

so e^ln .5 = 1/2

Answer 4

Let A = e^(-ln 2) = 1 / e^(ln 2)
let y = e^(ln 2)
ln y = ln 2.(ln e)
y = 2
A = 1 / 2