The standard equation for continuous exponential growth is:A = Pe^{kt}

...where "A" is the ending amount, "P" is the beginning amount, "k" is the growth (or decay) constant, and "t" is the time.

In your case, you have P = 123, and A = 440 at t = 5. Using this, you can solve for the constant k:

440 = 123e^{5k}

440/123 = e^{5k}

ln(440/123) = 5k

ln(440/123)/5 = k

(Growth/decay constants are often messy like this. Do NOT simplify or approximate with a truncated decimal! Round-off error can quickly become huge with logs and exponentials, so stay "exact" for as long as you can, trying only to round at the very end!)

To figure the number of cells at t = -2, plug in -2 for t and ln(440/123)/5 for k, along with 123 for P, and simplify. I get approximately 74. (Remember that you can't have fractions of cells, so you'll have to round to a whole number to get a REASONABLE answer; their answer is not physically reasonable.)

For the last part, plug in ln(440/123)/5 for k, 123 for P, and 10000 for A. Solve for the time t. (Decimals/fractions are okay here.)

Hope that helps!