Log equation $\log(2x-1) = -x+3$ with two non log values

Question

What is the correct approach to solving a log equation with more than one non log value? Please demonstrate using the following equation: $$\log(2x-1)=-x+3$$

Answer 1

For an equation that is similar to this one, you will typically need the Lambert $W$ function. I assume that your $\log$ is $\ln$.

$$\begin{align} \ln(2x-1)&=-x+3\\ 2x-1&=e^{-x+3}\\ (2x-1)e^{x-3}&=1\\ \left(x-\frac12\right)e^{x-3}&=\frac12&&\text{make coefficients of $x$ equal}\\ \left(x-\frac12\right)e^{x-1/2}&=\frac12e^{5/2}&&\text{make subtractions from $x$ equal}\\ W\left(\left(x-\frac12\right)e^{x-1/2}\right)&=W\left(\frac12e^{5/2}\right)\\ x-\frac12&=W\left(\frac12e^{5/2}\right)\\ x&=W\left(\frac12e^{5/2}\right)+\frac12\\ \end{align}$$ And $x\approx1.941304330652583916268007448815937227342114287347817503685\ldots$.

Answer 2

As alex jordan answered, there is an analytical solution involving Lambert function. If you cannot or do not want to use it, numerical methods should be required.

Consider the function $$f(x)=\log(2x-1)+x-3$$ and its first derivative $$f'(x)=\frac 2{2x-1}+1$$ It is always positive since, because of the logarithn, $x>\frac12$; so $f(x)$ is increasing and there is a single root.

By inspection $f(1)=-2$, $f(2)=\log(3)-1>0$. So, the root is beween $1$ and $2$.

Let us be very lazy and start Newton iterations at $x_0=\frac 32$ and apply the iterative scheme $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ This will generate as successive iterates $$\left( \begin{array}{cc} n & x_{n} \\ 1 & 1.903426410 \\ 2 & 1.941095404 \\ 3 & 1.941304324 \\ 4 & 1.941304331 \end{array} \right)$$