What is x in terms of a and b ?

Question

aˣ = x + b

Answer 1

Before I begin, I'll point out that the first respondent did not isolate the variable x, which is an interesting way to solve for x.

What you have here is a modification of the Lambert W-function. See: http://mathworld.wolfram.com/LambertW-Function.html

To make yours a bit more like the standard form, do as follows:

substitute -y=x.
this gives you (with some rearrangement):
(b-y)a^y = 1

substitute z=y-b, rearrange, and you get:
z a^z = 1/(a^b)

take the log of both sides, base a:
logbasea (z) + z = logbasea(-a^(-b))

change the logarithmic base and simplify:

ln(z) + z*ln(a) = ln(-a^(-b))

put everything back to the e power and reducing:

z e^(z*ln(a)) = -a^(-b)

one last substitution, w = z/ln(a) gives:

w e^w = - ln(a) * a^(-b)

Now we can finally use that page I linked to earlier. This is a standard, non-reducible transcendental equation in w. We cannot represent the solution symbolically, and cannot compute it without the values of a and b (although for very specific values of a and b, we may be able to solve it symbolically). Regardless, the best we can do is call this quantity:

w = PL( - ln(a) / a^b)

We then just go back through our substitutions to find:

x = -b - PL( -ln(a)/a^b ) / ln(a)

That's the best solution you're going to get, I'm afraid. There is no way to simplify this into any sort of elementary expression. I can give you a numerical approximation if you'd like, for specific values of a and b. For example, (a=2,b=3) gives x = -2.8625 or 2.44491 - this actually has two solutions.

That's right, the PL function is multivalued. If we, by typical convention, say PL(k) is some principle value based on a single solution to w e^w = k, then we are omitting extra solutions to this equation.

Man, that's really long, and I enjoyed doing it. I hope it helped!

Answer 2

x = aˣ - b