The strength, S, of any signal in a fiber-optic cable diminishes 15% for every 10 miles. Let S sub0 (So) be the strength at the location from which the signal is to be calculated. Let n, the horizontal axis, represent each 10 mile length. Find the formula for this decay-exponential function.

Graphically approximate the distance at which the signal has diminished to half its original strength.

Verify your answer using the formula you found above. :confused:

i found this page which was helpful
Exponential Decay (http://www.physics.uoguelph.ca/tutorials/exp/decay.html)

replace the N in their equations with S and the t with n to match your problem

first determine the decay constant using the equation \ln\left(\frac S S_0\right) = k n

in the case we have:
S = 0.15 S_0 \\
n=10


so k = ln(0.15)/10 = ln(0.15^0.1)

now put this into the other equation on that page:
S=S_0e^{kn} \\
S=S_0e^{\ln\left(0.15^{0.1}\right)n} \\
S=S_0e^{\ln\left(0.15^{0.1n}\right)} \\
S=S_0 0.15^{0.1n} \\
S\approx S_0 0.827197^n


Now graph this using a few key points, such as (n,S0) and (10,0.15*S0)
Look at the graph and estimate the n where S = 0.5 S0
for me, it looks to be about 3.6 or 3.7
Then solve using the equation above:
0.5 S_0=S_0 0.15^{0.1n} \\
0.5 = 0.15^{0.1n}\\
\log_{0.15} 0.5 = \log_{0.15}0.15^{0.1n}\\
\log_{0.15} 0.5 = 0.1n \\
n=10 * \log_{0.15} 0.5 \\
n=\log_{0.15} 0.5^{10} \\
n\approx 3.65368 \text {miles}

this is close to my estimate so it looks good.

let me know if you don't follow anything!

A=P(1-r)x