Question

How I can find this value 0.45=1-exp((-a/2))

Answer 1

We have
\[0.45= 1-e^{\frac{-a}{2}}\]
could you simplify it a bit?

Answer 2

\[\frac{ 9 }{ 20 }=1-\frac{ 1 }{ \sqrt{a} }\] like this ?

Answer 3

\[\frac{ 1 }{ \sqrt{e^a} }\]

Answer 4

I think we have like this
\[e^{\frac{-a}{2}}=0.55\]
Do you follow?

Answer 5

yes , i guess

Answer 6

how did you 0,55 ?

Answer 7

we had
\[0.45=1-e^{\frac{-a}{2}}\]
subtract both sides by 1
\[0.45-1=1-e^{\frac{-a}{2}}-1\]
now we get
\[-0.55=\cancel {1}-e^{\frac{-a}{2}}-\cancel {1}\]
\[-0.55=-e^{\frac{-a}{2}}\]
Multiply both sides by -1 and swap
\[e^{\frac{-a}{2}}=0.55\]

Do you follow?

Answer 8

i see

Answer 9

do you know logarithm?

Answer 10

used to know it but i dont recall it

Answer 11

okay natural log or log to the base e is defined as
\[\log_{e} x=y\]
\[\text{this implies}\\ x= e^{y}\]
do you follow?

Answer 12

yes i follow

Answer 13

okay let's apply log to the both sides in our expression
we have
\[e^\frac{-a}{2}=0.55\]
applying log to the base e
\[\log_e e^\frac{-a}{2}=\log_e 0.55\]

now per logarithmic property
\[\log_a a^b=b\]
so we get
\[\frac{-a}{2}=\log_e 0.55\]
Do you follow these steps?

Answer 14

k

Answer 15

now to find value of \[\log_e 0.55\] we need to refer log tables, do yo have one? or is this value given in question..

Answer 16

i don't a table but log 0.55 is like a negative numbers

Answer 17

log (0.55) =-.519274621

Answer 18

yes, it will be negative as 0.55 is less than e

We can use a calculator but it that allowed in your homework?

Answer 19

im allow to use a program call maple

Answer 20

but i still need to find the value of a

Answer 21

check it if there is log function there..

Answer 22

what i need to check?

Answer 23

please check this link..
http://www.maplesoft.com/support/help/maple/view.aspx?path=ln

Answer 24

ok

Answer 25

-.5978370008/ln(e)

Answer 26

okay so we have
\[-a/2=-0.5978\]

I think you can find a now...

Answer 27

is 1.195674002

Answer 28

yes, just simple math

Answer 29

k thx so you much for the help

Answer 30

No problem.