$\frac{dx}{dt} = kx \\ \implies \frac{dx}{x} = k dt \implies log_e x = kt + c$

Let initial number of bacterias be $x_0$

Thus, at $t=0, log_e x_0 = c$

Also, at $t=1, x = 2x_0 \implies log_e (2x_0) = k.1 + log_e x_0 \implies k = log_e 2$

Therefore, $log_e \frac{x}{x_0} = t log_e 2$

At $t=4, log_e \frac{x}{x_0} = 4 log_e 2 = log_e 2^4 = log_e 16 \\ \implies x_4 = 16 x_0$