Reversing equation with a logarithm and exponent

Question

This is my equation: $$ x=y^{3.333+(-1(0.5\times \log_{10}(10-y)))} $$ It will solve for x, with input of any y. I want to solve for y with input of any x.

Answer 1

I do not think that you could express analytically $y$ as an explict function of $x$ and only numerical methods can be used. Probably the simplest would be Newton which, starting from a "reasonable" guess $y_0$ will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$

For illustration purposes, let us consider the function you give (at least how I read it) $$f(y)=y^{\frac{10}{3}-\frac{\log (10-y)}{2 \log (10)}}-x$$ $$f'(y)=y^{\frac{10}{3}-\frac{\log (10-y)}{2 \log (10)}} \left(\frac{\frac{10}{3}-\frac{\log (10-y)}{2 \log (10)}}{y}+\frac{\log (y)}{2 (10-y) \log (10)}\right)$$ and consider the case where $x=1234$. A plot of the function shows that the solution is close to $8$; so, let us start iterating with $y_0=8$. The method then generates the following iterates : $9.03900$, $8.79341$, $8.74220$, $8.74060$ which is the solution for six significant figures. What you can notice is that we overshoot the solution because we started at a values where $f(y)f''(y) <0$. If, instead, we used $y_0=9$, the iterates would be $8.78027$ , $8.74150$, $8.74060$.

If we where looking for the $y$ corresponding to a "small" value of $x$, we could first approximate $f(y)$ by its Taylor expansion built at $y=0$. This would give $$f(y)\approx y^{17/6}-x$$ So, trying to solve the equation for $x=100$ would provide an estimate $y_0=10^{12/17}\approx 5.08$ and the iterates would then be $4.74450$, $4.71048$, $4.71016$ which is the solution for six significant figures.