# Solve for x?

Logarithms problems.

### Solve for x?

I've been a degreed mechanical engineer for 35 years and thought I could do algebra in my sleep. But I think I've met my Waterloo. I'm working with a phenomenon with exponential decay and I need to solve the following equation for x:

a log(1 - b/x) = c log(1 - d/x)

a, b, c & d are known parameters. Is this solvable to a general equation of x =_____?

Don
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### Re: Solve for x?

Okay, you have a log(1- b/x)= c log(1- d/x).
That is the same as log((1- b/x)^a)= log((1- d/x)^c).

Since log is a one-to-one function, we have (1- b/x)^a= (1- d/x)^c.

That's what you got, right? You can, I don't know if it helps, let y= b/x. Then x= b/y so d/x= dy/b and we can write the equation (1- y)^a= (1- dy/b)^c. That, at least, gets the unknown out of the denominator. You could also let z= 1- y so that y= 1- z. dy/b= d/b- dz/b so that 1- dy/b= 1- d/b-dz/b and the equation changes to z^a= (1- d/b-dz/b)^c but there is no general way to solve that,
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