Answer:

Step-by-step explanation:
Let's start by using change of base property:

So, for 

Now, using change of base for 

You can express
as:

Using reduction of power property:


Therefore:

As you can see the only difference between (1) and (2) is the coefficient
:
So:


EXPLANATION
Let's see the facts:
The scale is---> 14 milimeters ----> 5 meters
Width = 42 milimeters
Applying the unitary method, the actual width of the pool is:

The width of the pool is 15m
43.305 or in Mixed Number is 43 61/200 or in improper Fraction is 8661/200