Help req/d please not getting this

[ads114]

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.

 

[statdad]

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?

 

[ads114]

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

 

[statdad]

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

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

 

[ads114]

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?

 

[Mark44]

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?

 

[ads114]

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!

 

[Mark44]

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

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 = e^a.

 

[ads114]

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

 

[Bohrok]

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

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

 

[Mark44]

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

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

 

[ads114]

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.

 

[Mechdude]

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

 

[HallsofIvy]

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!

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 = e^a.

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

Essentially, then, you are telling us that you know that e^x 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)))= e^Ln(y/(1-y))= e^x-3. Now, what is the left side of that? What is e^{Ln(y/(1-y))? (Remembering that eLn(x)= x for any number x.)}

 

[ads114]

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, what's this all about? if you can't help without belittling me don't bother really its not worth the hassle.

 

[Mark44]

What you're taking as hostility coming from us might be our collective sense of frustration. Five people have jumped into 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.

 

[ads114]

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 don't 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.

 

[dharavsolanki]

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?