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Are my answers correct? Quadratic and logarithmic modeling

needalgebra

Member
Aug 13, 2013
45
This is what i got...

For option 1 I think it's y=( .5(x-20))^2+20 for x>20

For option 2 I think it's y=30+.5*ln|x-25| for x>25


Question:

develop equations that model the two replacement pack options

(option 1) quedratic model, monthly access fee \$20, included gigabytes 20, cost per additional gigabyte *see below (*)

(option 2) logarithmic model, monthly access fee \$30, included gigabytes 25, cost per additional gigabyte *see below (**)

(*) The charge option 1 will increase quadratically after 20 gigabytes with customers paying \$0.50.

(**) The charge option 2 will increase logarithmically in the form ƒ(x) = a + b 1n|x| after 25 gigabytes with customers paying \$0.50.
 
Last edited:

needalgebra

Member
Aug 13, 2013
45
Re: Are my answer correct?

I cant seem to set up correctly please help, its suppose to come up as normal text!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Are my answer correct?

I cant seem to set up correctly please help, its suppose to come up as normal text!
I had it fixed, but you unfixed it...:D

Okay, I fixed it again...dollar signs need a backslash before them, otherwise they are parsed as $\LaTeX$ tags.
 

needalgebra

Member
Aug 13, 2013
45
Re: Are my answer correct?

You are my hero. Thanks! (Sun)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Are my answer correct?

I am assuming in both cases, we want the first additional gigabyte to cost \$0.50.

So, for the quadratic model, you want:

\(\displaystyle f(x)=ax^2+b\)

where:

\(\displaystyle f(20)=a(20)^2+b=400a+b=20\)

\(\displaystyle f(21)=a(21)^2+b=441a+b=20.5\)

I would solve by elimination.

For the logarithmic model, you want:

\(\displaystyle f(x)=a+b\ln(x)\)

where:

\(\displaystyle f(25)=a+b\ln(25)=30\)

\(\displaystyle f(26)=a+b\ln(26)=30.5\)

I would use elimination here as well.
 

needalgebra

Member
Aug 13, 2013
45
Ok, is this right?


Quadratic model:

f(21) - f(20):
41a = 0.5
a = (1/2)/41 = 1/82
b = 620/41
f(x) = (1/82)x^2 + (620/41)


Logarithmic model:

f(26) - f(25):
bLn(26) - bLn(25) = 1/2
bLn(26/25) = 1/2
b = 1/(2Ln(26/25)) = 1/Ln(26/25)^2
a + Ln(25)/Ln(26/25)^2 = 30
a = 30 + Ln(16900)
f(x) = 30 + Ln(16900) + (Ln(x)/Ln(26/25)^2)
f(x) = 30 + 3Ln(25) + Ln(x)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The quadratic model is correct, but you have made an error in the logarithmic model. You have correctly found the value of $b$, but the error lies in the computation of $a$.

Here is the step where I see the error:

a + Ln(25)/Ln(26/25)^2 = 30

a = 30 + Ln(16900)

It appears you are saying:

\(\displaystyle \frac{\ln(25)}{2\ln\left(\frac{26}{25} \right)}=-\ln(16900)\)

How did you arrive at this?
 

needalgebra

Member
Aug 13, 2013
45
Honestly,

I had almost no clue how to finish up the Logarithmic expression..
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Honestly,

I had almost no clue how to finish up the Logarithmic expression..
You were doing well up to the point I cited. Instead of:

\(\displaystyle a=30+\ln(16900)\)

You want:

\(\displaystyle a=30-\frac{\ln(25)}{2\ln\left(\frac{26}{25} \right)}\)

Do you see why?
 

needalgebra

Member
Aug 13, 2013
45
no :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You stated (in a slightly different form):

\(\displaystyle a+\frac{\ln(25)}{2\ln\left(\frac{26}{25} \right)}=30\)

I have chosen to write \(\displaystyle 2\ln\left(\frac{26}{25} \right)\) where you have used the equivalent \(\displaystyle \ln\left(\left(\frac{26}{25} \right)^2 \right)\).

So, to solve for $a$, you need to subtract \(\displaystyle \frac{\ln(25)}{2\ln\left(\frac{26}{25} \right)}\) from both sides to obtain:

\(\displaystyle a=30-\frac{\ln(25)}{2\ln\left(\frac{26}{25} \right)}\)

It's just like if you had:

\(\displaystyle x+y=z\)

and you solved for $x$ to get:

\(\displaystyle x=z-y\)

Does this make sense?
 

needalgebra

Member
Aug 13, 2013
45
yeah i think i got it.

is this correct?

b=1/(2*log(26)-2*log(25))
b=12.74836584551409
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is closer:

\(\displaystyle b\approx12.7483658455141791503275149034786729902710597570291566009838101858237\)

But...why use a decimal approximation when you can use the exact value? (Nerd)

I would use:

\(\displaystyle b=\frac{1}{2\ln\left(\frac{26}{25} \right)}\)
 

needalgebra

Member
Aug 13, 2013
45
I've been working on several different math tasks almost non-stop for the past 10 hours or so, i have a ton of work to do that i didnt know off. All of this is due in 3 days.. so you're going to have to excuse me if i dont make sense sometimes (feeling drowsy) - besides most of this work is stuff i havent touched in a real long time. :D haha. what else would i have to do to the Quadratic model? how would i finish off the logarithmic model?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You completed the quadratic model correctly, and you have the values of $a$ and $b$ for the logarithmic model as well.

Quadratic model:

\(\displaystyle f(x)=\frac{1}{82}x^2+\frac{620}{41}\) where \(\displaystyle 20\le x\)

Logarithmic model:

\(\displaystyle f(x)=\left(30-\frac{1}{2\ln\left(\frac{26}{25} \right)} \right)+\frac{1}{2\ln\left(\frac{26}{25} \right)}\ln(x)\) where \(\displaystyle 30\le x\)
 

needalgebra

Member
Aug 13, 2013
45
omg.... i feel so dumb right now. haha

Thanks for your help! Is there anyway of giving you some kind of reward on here, like givin you 500000/10 for being an amazing helper? lol if i knew you in real life i'd take you to starbucks for a cup of coffee :D (Ninja)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hey, if you really want to give me a reward, tell your classmates about us and encourage them to register. :D

But, we enjoy helping here, and if you gained a bit of understanding from your time here, then our goal here at MHB has been met. I look forward to seeing you around!
 

needalgebra

Member
Aug 13, 2013
45
I actually study online, but i will definitely let my online - classmates know!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

needalgebra

Member
Aug 13, 2013
45
how would i graph the logarithm and quadratic models? having a little trouble here...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Use the command:

piecewise[{{20,0<=x<=20},{x^2/82+620/41,20<x}}],piecewise[{{30,0<=x<=25},{(30-log(25)/(2log(26/25)))+1/(2log(26/25))log(x),25<x}}] where x=0 to 50

at W|A.
 

needalgebra

Member
Aug 13, 2013
45
i've been trying but cant seem to get it.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

needalgebra

Member
Aug 13, 2013
45
How do i get the intersection points of the logarithm and quadratic models?
 

needalgebra

Member
Aug 13, 2013
45
ok so for x i got :

x^2/82+620/41=
(log(26/25)*log(x))/2-log(26/25)/2+30

x=34.98726

cant seem to get y.