How to find analytically the third root (-ve) of $x^{2}=2^{x}$ other than 2 and 4?
Does differentiation of the equation make a sensible way?
I tried with $\log_a$ for different $a$' s. But I couldn't find the root.
How to find analytically the third root (-ve) of $x^{2}=2^{x}$ other than 2 and 4?
Does differentiation of the equation make a sensible way?
I tried with $\log_a$ for different $a$' s. But I couldn't find the root.
Looking for the negative solution, we get $$ x^2=2^x\\ x^2=e^{x\ln 2}\\ -x=e^{x/2\cdot\ln 2}\\ -xe^{-x/2\cdot\ln 2}=1\\ -\frac x2\ln 2e^{-x/2\cdot\ln 2}=\frac12\ln 2\\ -\frac x2\ln 2=W\left(\frac{\ln 2}2\right)\\ x=-\frac{2W\left(\frac{\ln 2}2\right)}{\ln 2}\approx-0.766665 $$ where the sign on the left-hand side in line 3 comes from knowing that $x$ is negative, and $W$ is the Lambert $W$ function. Which is to say, $W(1/2\cdot\ln 2)$ is the solution to $$ ye^y=\frac{\ln 2}2 $$