Isolating variable in exponents in a linear combination of exponential functions

Question

I'm doing a physics problem that's asking me to compare the maximum speeds of a simple harmonic oscillator versus one immersed in a fluid that's leading to overdamping. I am at a point where I have the velocity solution: $$\dot x(t) = \frac{-\omega^{2}x_0}{2\Omega}[e^{-(\gamma - \Omega) t} - e^{-(\gamma + \Omega)t}]$$ Where $$\Omega = \sqrt{\gamma^{2} - \omega^2}$$ All Greek letters are constants. From here I'd like to differentiate and set the second derivative equal to zero then solve for $t$ to determine the time in which the velocity is at a maximum then plug that back in to the equation for the velocity to determine the maximum velocity. I'm having trouble solving for $t$ and would like some assistance if possible. The second derivative is then: $$\ddot x(t) = \frac{-\omega^{2}x_0}{2\Omega}[-(\gamma - \Omega)e^{-(\gamma - \Omega) t} + (\gamma + \Omega)e^{-(\gamma + \Omega)t}] = 0$$ From here I know if I want to solve for $t$ then I'll need to employ the natural logarithm to "pull" the $t$ from the exponent down to the base however I'm confused on a few fronts, namely:

1. Would I be applying the natural log to the entire function or could I selectively apply it only to the 2 exponential functions?
2. Furthermore, would I need to apply the natural log to both sides of the equation (as if I was multiplying one side by a constant?). Then I would get a $-\infty$ on the RHS and that doesn't seem right.

I appreciate any help in elucidating the way in which one would work with logarithms to help isolate the variable in question here.

Answer 1

Would I be applying the natural log to the entire function or could I selectively apply it only to the 2 exponential functions?

The log of a sum does not help much. But you can firstly factor out $e^{-(\gamma - \Omega)t}$

$$e^{-(\gamma - \Omega) t}\cdot\frac{-\omega^{2}x_0}{2\Omega}[-(\gamma - \Omega) + (\gamma + \Omega)e^{-2\Omega t}] = 0$$

Now we use the following rule: A product is zero, if at least one of the factors is zero.

With some assumption we can say that the factors $e^{-(\gamma - \Omega) t}$ and $\frac{-\omega^{2}x_0}{2\Omega}$ cannot be zero. Thus the bracket must be zero.

$-(\gamma - \Omega) + (\gamma + \Omega)e^{-2\Omega t}=0$

$ (\gamma + \Omega)e^{-2\Omega t}=(\gamma - \Omega) $

$ e^{-2\Omega t}=\frac{\gamma - \Omega}{\gamma + \Omega} $

Now you can take logs.