# Help req/d please not getting this

1. Oct 29, 2009

help req/d please not getting this!!!!!!!!!!!!!

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 + In( y/1-y)

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

can anyone give me a hint

thankyou.

2. Oct 29, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

You have an equation, a value of x, and need the value of y. What do you need to do with the equation to solve for y?

3. Oct 30, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

hi there obviously i need to transform this equation for y?

4. Oct 30, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

Yes - what you have it as now gives x as a function of y - you need to solve to get y as a function of x

5. Oct 30, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

ok,i see where you are going with this!
how do i transform the equation to "y" only,especially with the "In" ?
suspect i need to multiply the brackets through by the "In" function which will then simplify it?
hmn,not sure, this is confusing again,"y" is top and bottom! or should I put in the values 1st for "y" that may show it more clearly perhaps?

6. Oct 30, 2009

### Staff: Mentor

Re: help req/d please not getting this!!!!!!!!!!!!!

First, add -3 to both sides of your equation. Then you're going to need the inverse of the Ln function (there is no In function). Do you know which function is the inverse of Ln?

7. Oct 30, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

hello mark,thankyou for assisting me,inverse of the Ln function is the "e" function but adding -3 will get rid of the +3 on the Lhs but how does this help as its now -3 on the rhs!

8. Oct 30, 2009

### Staff: Mentor

Re: help req/d please not getting this!!!!!!!!!!!!!

Adding -3 to both sides gives this:
x - 3 = Ln(y/(1 - y))

The idea you need to use is that a = Ln b <==> b = ea.

9. Oct 30, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

apologies mark.you have lost me there,think you see my problem, where did b and a come from!

10. Oct 30, 2009

### Bohrok

Re: help req/d please not getting this!!!!!!!!!!!!!

If a = Ln(b), then b = ea for any real number a and b>0.

11. Oct 31, 2009

### Staff: Mentor

Re: help req/d please not getting this!!!!!!!!!!!!!

I'm telling you what to do next without working the problem for you.

12. Oct 31, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

mark suspect i irritated you there. i do not want you answer the question for me,that is up to me.i am just not understanding it.

13. Oct 31, 2009

### Mechdude

Re: help req/d please not getting this!!!!!!!!!!!!!

Ads111 this is what mark wants you to see: $$\frac {y}{1-y} = e^{x-3}$$

14. Oct 31, 2009

### HallsofIvy

Staff Emeritus
Re: help req/d please not getting this!!!!!!!!!!!!!

Essentially, then, you are telling us that you know that ex is the "inverse" of the Ln(x) function but you don't know what that means!

If you f-1(x) is the inverse to f(x), that means that f(f-1(x))= x and that f-1(f(x))= x. That is the standard way of solving equation.

Once you have Ln(y/(1-y))= x- 3, take exp() (another name for e()) of both sides. exp(Ln(y/(1-y)))= eLn(y/(1-y))= ex-3. Now, what is the left side of that? What is eLn(y/(1-y))? (Remembering that eLn(x)= x for any number x.)

15. Oct 31, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

Re hallsofivy........... to be honest feel i'm getting a lot of hostility here, i asked for help with an equation i did'nt understand, i don't need a not a lot of smart comments, whats this all about? if you can't help without belittling me don't bother really its not worth the hassle.

16. Oct 31, 2009

### Staff: Mentor

Re: help req/d please not getting this!!!!!!!!!!!!!

What you're taking as hostility coming from us might be our collective sense of frustration. Five people have jumped in to give you a push in the right direction, and have nearly worked this problem for you. This problem is fairly straightforward if you understand the basic idea about a function and its inverse, and in particular, the relationship between the natural log function and the exponential function, both of which we have provided. If you don't understand this concept, don't just say you don't understand, ask a specific question about it -- don't just say you don't understand.

17. Nov 1, 2009

Re: help req/d please not getting this!!!!!!!!!!!!!

Hello mark,many thanks for your words of encouragement.
So we are clear, I am not knowledgeable about the functions of logs,natural or otherwise,the exponential function and Ln function are also difficult to understand.
When i am saying "i do not understand" it is actually because i dont understand the equation or indeed how to go about transforming it for "y" or anything else associated with it
If yourself is willing or anyone else able to explain or simplify it in some manner that will be super.
To clarify to everyone, this is not my attempt to have this answered, that is up to me to do but please remember maths is not for everyone.

18. Nov 1, 2009

### dharavsolanki

Re: help req/d please not getting this!!!!!!!!!!!!!

Exactly. You got the situation damn right.

Listen, you took the 3 on RHS and made it a -3 on the LHS. The idea is NOT GETTING RID OF 3. The idea is that e to the power of ln of anything is that thing.

e(ln a) = a

To explain this, you were given another example which used a and b. a and b cold be anything - it was jst a random example - they didn't come in from anywhere.

The point till now is, why do we require -3 on LHS instead of +3 on RHS?

After you take in the inverse function's relations - the y comes out of the natural log. Isn't that simple?