• Nov 5th 2009, 12:10 AM
hi guys can someone assist with this one please before i pull any more hair out.

the age of a machine,x, in years is related to the probability of breakdown,y, by the formula

x=3 + ln( y/1-y)

determine the probability of breakdown for 1,3,10 years!

i am fairly sure 1st i take away the 3 from both sides thus leaving x-3=ln(y/1-y).

then inverse of ln makes y/1-y=e^x-3?

what do you think any help appreciated,logs,powers etc not my strong point!

ps. hope this is the correct section to be in.

thankyou
• Nov 5th 2009, 12:52 AM
Quote:

the age of a machine,x, in years is related to the probability of breakdown,y, by the formula

x=3 + ln( y/1-y)

determine the probability of breakdown for 1,3,10 years!

i am fairly sure 1st i take away the 3 from both sides thus leaving x-3=ln(y/1-y).

then inverse of ln makes y/1-y=e^x-3?
yes, now multiply both sides by 1-y, and move the ye^x-3 to the left and factorise out the y and then divide both sides by e^x-3

y= (1-y) (e^x-3)

y=e^x-3 - y e^x-3

y + y e^x-3=e^x-3

y(1 + e^x-3)=e^x-3

y=e^x-3/(1 + e^x-3)

now you got y by itself.

sub in those values of x to get probabilty.

it seems correct but correct me if im wrong
• Nov 5th 2009, 01:04 AM
ah of course! how stupid of me.

i was actually trying to cross multiply which didnt leave y on its own ,many thanks for the very quick reply.(Hi)
• Nov 5th 2009, 01:11 AM