HINT:#
The question asks us to determine the relative distances that the two bulbs should be from the observer for the bulbs to have the same apparent brigthness.
HINT:#
We know the Wattage of the two bulbs, and from the previous explanation, we have a basic idea of how the wattage of a bulb affects the brightness of the bulb.
P1 = 25 Watts, and P2 = 100 Watts
HINT:#
We know that light is radiating from the bulbs in all directions at a constant rate. To see The invserse square law at work, let's consider the light leaving the bulbs at one instant. Since the radiation emitted by the bulb is related to the amount of electricity being converted into light, the higher wattage bulb emits more photons per instant. If we think of all of these photons as spread over the surface of a sphere the size of the bulb, the surface of the higher wattage bulb will have a more dense coating of photons on it. Let's also recall that the Surface Area of a sphere is given by
A = 4Πr2
If we say that the total number of photons emitted per instant by the bulb is equal to P, the number of photons per unit area of the sphere is given by
Φ = P/A = P/4Πr2
Now let's consider what happens when the photons begin to move away from the inital spherical surface. Since the radiation is being emitted in all directions, we can think of the photons as moving away from the light source, so, in a direction 'normal' to the surface of the sphere. We can think of the photons as pushing the surface of the sphere away from the center, or as increasing the radius of the sphere at the speed of light! We describe the number of photons moving through a unit square of the sphere a distance r from the source with the term flux, denoted by the Greek symbol Phi, Φ.

If we P1 is the power of the weaker bulb and P2, that of the stronger bulb, and we want to observe the same brightness from both bulbs, we want the flux from both bulbs to be the same, and r1 and r2 are the distances between the first and second bulb repsectively,
P1/4Πr12 = P2/4Πr22
P1/r12 = P2/r22
r12/r22 = P1/P2
r1/r2 = (P1/P2)1/2
HINT:#
P1 = 25 Watts
P2 = 100 Watts
HINT:#
Let's use unit analysis:
Since both sides of the equation are ratios of quantities with the same units, the units will cancel and our answer can with be expressed as a dimensionless factor
r1/r2 = distance factor
 or as a known distance multipled by the factor, r1 = r2*factor.
No conversions will be necessary.
HINT:#
r1/r2 = (25 Watts/100 Watts)1/2 = 0.5.
 So, the weaker bulb must be half as far from the observer as the more powerful bulb to observe the same brightness.

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