How to solve for t in a exponential polynomial function

Question

How to simplify and solve for $t$ in functions of the following form:

$\ 46080e^{-600t} - 576000e^{-1500t} + 737280e^{-2400t}= 0 $

Answer 1

Given expression can be written as $$10\cdot e^{-600t}\cdot (73728e^{-1800t} -57600e^{-900t} + 4608) = 0$$

If $t$ is finite $e^{-600t} \neq 0$

Now, we put $x= e^{-900t}$.

Thus the expression reduces to

$$73728x^2 - 57600x +4608=0$$

We can find the value of $x$ by Sridharacharya's formula.

The roots of a quadratic equation $ax^2+bx+c$ are $$\boxed{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

$\Large x=\frac{57600 \pm \sqrt{56700^2-4\cdot 73728 \cdot 4608}}{2(73728)}$

On Calculating
$x=0.69$ or $x=0.09$

Hence,
$e^{-900t}=0.69$ or $e^{-900t}=0.09$

Taking natural logarithm on either sides
$-900t = \ln 0.69$ or $-900t=\ln 0.09$

Hence, $\LARGE \boxed{ t= \frac{-\ln 0.69}{900}}$ or $\LARGE \boxed{t= \frac{-\ln 0.09}{900}}$

Answer 2

If you multiply by $e^{2400 t}$ it looks like $$ax^2+bx+c=0$$ where $x=e^{900t}$.