How do I accurately model and calculate costs for different data plans?

[needalgebra]

whaaaaat how did you get that?

 

[Deveno]

Substituting in for \(\displaystyle x\) obviously shows MarkFL's answer is correct...however, it does not directly give you \(\displaystyle a\) and \(\displaystyle b\).

To do this, re-write:

\(\displaystyle f(x) = 30 + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln\left( \frac{x}{25}\right)\)

\(\displaystyle = 30 + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln(x) - \frac{30\ln(25)}{\ln\left(\frac{12}{5}\right)}\)

\(\displaystyle = 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln(x)\)

which tells you:

\(\displaystyle a = 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) \approx -80.302\)

\(\displaystyle b = \frac{30}{\ln\left(\frac{12}{5}\right)} \approx 34.267\)**********

Solving the linear system:

\(\displaystyle a + b\ln(25) = 30\)
\(\displaystyle a + b\ln(60) = 60\), I obtain (subtracting the top equation from the bottom):

\(\displaystyle b(\ln(60) - \ln(25)) = 30\)

\(\displaystyle b = \frac{30}{\ln(60) - \ln(25)} = \frac{30}{\ln\left( \frac{12}{5}\right)}\)

Substituting this in the first equation:

\(\displaystyle a = 30 - b\ln(25) = 30 - \left( \frac{30}{\ln\left( \frac{12}{5}\right)} \right)\ln(25)\)

\(\displaystyle = 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) \)

which also agrees with MarkFL's answer.

 

[needalgebra]

ohhh ok.

I appreciate all of the help I am getting, thanks guys. I am actually done with this topic, i just want to know everything exactly how it is.

Thank you! How would i put that into wolfram for the graph?

and how would i do it without the included gigabytes, and the graph for that one?

 

[needalgebra]

any help guys please? :)