Solve $\frac{120000}{1+48e^{-0.015t}}=24e^{0.055t}$

Question

This is a challenge problem for Algebra 2 class

Solve $$\frac{120000}{1+48e^{-0.015t}}=24e^{0.055t}$$

Answer 1

There's no easy to solve it. However, you can try this:

$$120000=24e^{0.055t}(1+48e^{-0.015t})$$

$$120000=24e^{0.055t}+1152e^{0.04t}$$

Let $x=e^{0.005t}$

$$120000=24x^{11}+1152x^8$$

$$24x^{11}+1152x^8-120000=0$$

So now you find the positive root ($x\approx 1.76337$)

And $t=200\ln(x)\approx113.446$

Side note: Whoever made this "challenge problem" ... why