Question

Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-optic cable, its intensity decreases with the increase in the length of the cable. If 1500 lumens of light enters the cable, the intensity of light decreases by 3.4% per meter of the cable. Write a function f(x) to represent the intensity of light, in lumens, when it has passed through x meters of the cable.

Answer 1

After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it.  So only  (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter,  96.6%  of what entered it emerges from it, and
that's  96.6%  of  96.6%  of the original signal that entered the beginning
of the fiber.

==>  After 2 meters, the intensity has dwindled to  (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==>  After  'x'  meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
                                                     (1,500) · (0.966)^x
lumens of light remaining.
 
=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say  "15dB per 100 meters".

What does that mean ?    Break it down:  15dB in 100 meters is 0.15dB per meter.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is  3.4% .  We applied
super high mathematics to that and calculated that  96.6% remains, or  0.966.

Look at this  ==>      10 log(0.966) =  -0.15    <==  loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in  'x'  meters of
fiber cable, doesn't have to mess with raising numbers to powers.  All he has to
do is say ...

--  0.15 dB loss per meter

--  'x' meters of cable

--  0.15x dB of loss.

If  'x' happens to be, say,  72 meters, then the loss is  (72) (0.15) = 10.8 dB .

and  10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083  =  8.3%  <== That's how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.