Here we have the typical “automatic dark detector” circuit:
Let’s assume LDR=5Kohm with light, and LDR=2Mohm in darkness, the battery is 12V instead of 9V, and the transistor has a forward voltage Vbe=0.7V.
With light, per the voltage divider formula, we have Vbase=0.57V and the transistor is cut off.
In darkness, the formula gives Vbase=11.42V. Since this is much higher than 0.7V. Why does the transistor not burn?