Solving equations with log

Question

Is there any easy way (no numerical methods, just a straightforward way to do it by hand) to solve an equation like:

\begin{equation}\frac{-34450}{x} - 2.01\ln(x) + 33.74 = 0\end{equation}

It is a step in a bigger problem that I'm trying to solve and the solution that was given to me just says "just solve the equation" as if that step was naturally trivial, but I can't see it.

Answer 1

A completely nice and analytical solution to this equation doesn't exist, the best attempt uses the Lambert W function, which is the inverse of $$xe^x$$ so $$W(xe^x)=x$$ Before applying this we need to move some things around: substitute $x=e^u$ and obtain$$-34450e^{-u}-2.01u+33.74=0$$ Let's call the constants a,b,c all positive real values and substituting $y=c-bu$ gives $$-ae^{\frac{y-c}{b}}+y=0$$ rearanging gives $$-\frac{a}{b}e^{-\frac{c}{b}}=-\frac{y}{b}e^{-\frac{y}{b}}$$ and after applying Lambert W we get $$y=-bW(-\frac{a}{b}e^{-\frac{c}{b}})$$ and resubstituting $$x=\exp(\frac{c}{b}+W(-\frac{a}{b}e^{-\frac{c}{b}}))$$

Answer 2

This may not be the answer you are seeking, but if you are seeking an approximate answer, a heuristic approach si the iterative one.

Your equation has the form

$$ -\frac{a}{x} - b \ln(x) + c = 0 $$

which can be written as $x = f(x)$ where

$$ f(x) = \frac{a}{c- b \ln(x)}$$

One begins by setting the logarithm to unity,

$$ x_0 = \frac{a}{c - b}$$

and then iterates $x_{n+1} = f (x_{n})$ a few times. If the value is converging it approaches the correct one.

Of course this is far from an exact solution but it doesn't need a calculator.

Answer 3

Without any special function

Working with whole numbers, you look for the zero of function $$f(x)=-\frac{3445000}{x}-201 \log (x)+3374$$ which,if $x>1$, is smaller than $$g(x)=-\frac{3445000}{x}+3374$$ So, we have a lower bound $$x_0=\frac{1722500}{1687}=1021.04\cdots$$ Now, use the tangent at this point (this is Newton method) that is to say, write $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ to successively obtain $$\left( \begin{array}{cc} n & x_n \\ 0 & 1021.04 \\ 1 & 1469.18 \\ 2 & 1768.40 \\ 3 & 1846.47 \\ 4 & 1850.30 \\ \end{array} \right)$$