Solving logarithm with variable in base and exponent for x

Question

$$ log_{1+\frac{0.0231}{x}}{5x} = 10/9$$

What is the solution to this equation and how would I approach the problem? For the purposes of the question, algebraic manipulation with log and exponential rules would be a preferred solution.

Answer 1

Assuming that the equation is $$\log_{1+\frac{0.0231}{x}}({5x}) =\frac {10}9$$ consider that you look for the zero of function $$f(x)=\log_{1+\frac{0.0231}{x}}({5x}) -\frac {10}9=\frac{\log (5 x)}{\log \left(1+\frac{0.0231}{x}\right)}-\frac {10}9$$ or $$g(x)=9\log (5 x)-10 \log \left(1+\frac{0.0231}{x}\right)$$ Compute the derivative and set it equal to $0$; this would give $$g'(x_*)=0 \implies x_*=-\frac{1463}{30000}< 0$$ So, since $x >0$, $g(x)$ is an increasing function. Observe that $$g\left(\frac{1}{5}\right)=-10 \log \left(\frac{2231}{2000}\right)\approx -1.09303$$ $$g\left(\frac{2}{5}\right)=9 \log (2)-10 \log \left(\frac{4231}{4000}\right)\approx5.67688$$so the solution is just above $\frac{1}{5}$. Then, start Newton method to get the follosing iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.2000000000 \\ 1 & 0.2217834116 \\ 2 & 0.2231295997 \\ 3 & 0.2231340463 \end{array} \right)$$

Edit

If we consider the more general case of the equation $$\log_{1+\frac{a}{x}}({bx}) =c$$ rewriting as above $$g(x)=\log (b x)-c \log \left(1+\frac{a}{x}\right)$$ and assuming that $a$ is small, we could expand the second logarithm as a very truncated Taylor series and get $$g(x) \sim \log (b x)-\frac{a c}{x}$$ and its approximate zero would be given by $$x \sim \frac{a c}{W(a b c)}$$ where appears Lambert function.

Now, if $0 \leq t \leq 1$, we have, as an approximation, $W(t) \sim \frac{t}{t+1}$ and then $$x \sim a c+\frac{1}{b}$$ Applied to the present case, this would give $$x \sim \frac{677}{3000}\approx 0.225667$$ which is not bad.