How to solve $q= \frac{\ln{n}}{\ln{b} + \ln{q}+\ln\ln{n}}$

Question

I have come across this equation recently. All the variables are positive and real too.

$$q= \frac{\ln{n}}{\ln{b} + \ln{q}+\ln\ln{n}}.$$

Under what conditions can this be solved for $q$?

Answer 1

Let $q = e^x$, so that $\ln q = x.$ Also, put $A = \ln{n},$ $B = \ln{b},$ and $C = \ln{\ln{n}}.$ Then your equation takes the form

$$e^x \; = \; \frac{A}{B + x + C}$$

Multiplying both sides by $B + x + C$ gives

$$ e^x \left(B + x + C \right) \; = \; A$$

Now make the variable change $u = B + x + C,$ so that $x = u - B - C.$ Then we get

$$e^{u - B - C} \cdot u \; = \; A$$

Multiplying both sides by $e^{B+C}$ gives

$$e^u \cdot u \; = \; Ae^{B+C}$$

Now we can express the solution in terms of the Lambert function:

$$ u = W\left(Ae^{B+C}\right)$$

Hence, recalling that $u = B + x + C$ and $q = e^x$, we get

$$ B + x + C \; = \; W\left(Ae^{B+C}\right)$$

$$ x \; = \; W\left(Ae^{B+C}\right) \; - \; B - C$$

$$ e^x \;\; = \;\; \exp\left[W\left(Ae^{B+C}\right) - \; B - C \right]$$

$$ q \;\; = \;\; \exp\left[W\left(Ae^{B+C}\right) \; - \; B - C \right]$$

$$ q \;\; = \;\; \exp\left[W\left(\left(\ln{n}\right)e^{\ln{b} + \ln{\ln{n}}}\right) \; - \; \ln{b} - \ln{\ln{n}} \right]$$

(a few minutes later) Using $e^{\ln{b} + \ln{\ln{n}}} \; = \; e^{\ln{b}} \cdot e^{\ln{\ln{n}}} \; = \; b\ln{n},$ we get

$$ q \;\; = \;\; \exp\left[W\left(b\left(\ln{n}\right)^2\right) \; - \; \ln{b} - \ln{\ln{n}} \right]$$