higher maths . . . exponential and logarithmic functions . .?

Question

For the formula F(m) = 20 e ^-m

a) find F(0)

Ive done this part and got the correct answer, 20 but i need someone to show me how to do b)

b) Find m such that F(m) = 1/2 F(0)

Apparently the answer is 0.693


All help is greatly appreciated =] x

Answer 1

You're definitely done a) correctly.

To do b):

F(m) = (1/2)F(0)

But F(0) = 20, so

F(m) = (1/2)(20) = 10

But F(m) = 20e^(-m), so we equate these two.

10 = 20e^(-m)

Divide both sides by 20,

1/2 = e^(-m)

Put this in exponential form.

log[base e](1/2) = -m

But remember that log[base e] is equal to ln, so

ln(1/2) = -m
m = -ln(1/2)

Answer 2

F(m) = 1/2F(0) = (1/2)(20) = 10, from part a

So 10 = 20 e^-m
1/2 = e^-m
Take the natural logarithm of both sides:
ln(1/2) = -m
so m = - ln(1/2) or ln(2) which is approximately 0.693

Answer 3

F(m) = 20 e ^(-m)

a) F(0) = 20 e^(0) = 20
[Note: -0 is still 0.]

b) F(m) = 1/2 F(0)
20 e^(-m) = (1/2)(20)
20 e^(-m) = 10
e^(-m) = 1/2
-m = ln(1/2)
-m = 0.693 (to 3 sig fig)

Answer 4

a) f(0)=? como F(m) =20e^(-m) =20*1/e^m
Logo F(0) = 20*1/e^0 = 20*1/1 = 20*1 =20

b) 1/2*F(0)=F(m)
Logo 1/2*20=20*e^(-m) ==>1/2=e^(-m)==> ln(1/2) = ln{e^(-m)}
Portanto ln(1/2) = -m*ln e==> ln(1/2)= -m==> m= -ln(1/2)
como m= -1*ln(1/2) ==>m = ln(1/2)^(-1) ==>
m = ln(2/1) ==> m = ln2

Answer 5

You know F(0) = 20 so ½F(0) = 10

So the problem is

10 = 20 exp(-m)

First divide by 20

½ = exp(-m)

take the natural log of both sides:
ln(½) = ln(exp(-m))
-0.693 = -m
m = 0.693

Answer 6

a)
F(m) = 20.e^(-m)
F(0) = 20 . e^(0)
F(0) = 20 (any number to power zero = 1)

b)
10 = 20e^(-m)
1 / e^(-m) = 2
e^(m) = 2
m ln e = ln 2
m = ln 2 (ln e = 1)
m = 0.693

Fae a Fifer!

Answer 7

it loses 1/5 or .2 every month it keeps 4/5 or .8 every month H = 25 * .8^3 H = 12.8 cm 25 * .8^x > 16 .8^x > 16/25 .8^x > .64 x > ln .64 / ln .8 x > 2 any time after 2 months