Why a transcendental equation can not be analytically evaluated I'm reading this book in Classical Mechanics and they derive an equation for the time a projectile takes to reach the ground once is fired (accounting for air resistance):
$$T=\frac{kV+g}{gk}(1-e^{-kT})$$
I do not have any questions on how they derive the equation or about the equation itself.My question is on what they say after formulating such equation:
"This is a transcendental equation, and therefore we cannot obtain an analytic expression for T."
I know transcendental equations are logarithm functions, trig functions or exponential functions, but I still do not understand when they say "you cannot obtain an analytical expression".
What do they mean ?

Typically, equations which involve mixtures of polynomial, trigonometric, exponential terms do not show explicit solutions in terms of elementary functions (the solution of $x=\cos(x)$ is the simplest example I have in mind).
However, you will be pleased (I hope !) to know that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has solutions which express in terms of Lambert function $W(x)$ which is defined by $x=W(x)e^{W(x)}$. As metacompactness commented, the solution of the equation you posted is given by $$T=a+\frac{W\left(-a\, k\, e^{-a k}\right)}{k}$$ where $a=\frac{kV+g}{gk}$.
You will learn about Lambert function which is a beautiful one (Lambert and Euler worked together). In the real domain, $W(x)$ requires $x \geq -\frac 1e$. So, for your case with $a\, k\, e^{-a k} \leq \frac 1e$, no problem since this always holds.
