One can use Taylor series of the log or exp function to get the result that $x = 1$. I was wondering if there is any other simple solutions.
Thanks a lot!
[Math] Solving $\log(x) = x-1$
algebra-precalculuslogarithms
algebra-precalculuslogarithms
One can use Taylor series of the log or exp function to get the result that $x = 1$. I was wondering if there is any other simple solutions.
Thanks a lot!
Best Answer
The function $f(x) = x - \log x$ has derivative $f'(x) = 1 - \frac1x$ which is negative for $x<1$ and positive for $x>1$. Therefore, $f$ is strictly decreasing for $x<1$ and strictly increasing for $x>1$. Since $f(1)=1$ we have $f(x)>1$ for $x<1$ and $x>1$. Therefore, $x=1$ is the only solution to $f(x)=1$.