Are my answers correct? Quadratic and logarithmic modeling

[needalgebra]

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>25Question:

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.

 

[needalgebra]

Re: Are my answer correct?

I can't seem to set up correctly please help, its suppose to come up as normal text!

 

[MarkFL]

Re: Are my answer correct?

I can't 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]

Re: Are my answer correct?

You are my hero. Thanks! (Sun)

 

[MarkFL]

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]

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]

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]

Honestly,

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

 

[MarkFL]

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]

no

 

[MarkFL]

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]

yeah i think i got it.

is this correct?

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

 

[MarkFL]

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]

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 don't make sense sometimes (feeling drowsy) - besides most of this work is stuff i haven't 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]

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]

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]

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]

I actually study online, but i will definitely let my online - classmates know!

 

[MarkFL]

I actually study online, but i will definitely let my online - classmates know!

Good deal! (Yes)

I have created a new topic for your next questions...we prefer new topics for new questions so that our topics do not become convoluted and hard to follow.

The new topic can be found here:

http://mathhelpboards.com/pre-algebra-algebra-2/mathematical-modeling-6006.html

 

[needalgebra]

how would i graph the logarithm and quadratic models? having a little trouble here...

 

[MarkFL]

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]

i've been trying but can't seem to get it.

 

[MarkFL]

Just copy and paste the command I gave above into the text entry field at:

Wolfram|Alpha: Computational Knowledge Engine

 

[needalgebra]

How do i get the intersection points of the logarithm and quadratic models?

 

[needalgebra]

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.

 

[MarkFL]

You don't seem to quite have the correct equation. To solve this would require using a numeric root finding technique. By the graph it looks like $x$ is closer to 42.

 

[needalgebra]

i have no diea how to do that, do you mind showing me how?

 

[MarkFL]

For the functions involved, Newton's method would be extremely tedious, and unless you understand differential calculus it would not be worth my effort to crank it out manually. For some reason, W|A will only give the root in which we are not interested.

I recommend you put the functions into your graphing calculator, and zoom in on the point of intersection until you have the desired accuracy.

 

[needalgebra]

I got this...

36.9999 = y
42 = x

 

[MarkFL]

\(\displaystyle (x,y)\approx(42,37)\)

seems pretty close to me, particularly if we can be satisfied with integral coordinates.

I "tricked" W|A into giving closer approximations:

\(\displaystyle (x,y)\approx(41.97171992699835,36.60518626378525)\)

 

[needalgebra]

Thanks!

I have one more graph to do.

I have to do it without the included gigabytes.

Here is what i got:

Quadratic model : f(x) 1/82 x2

domain: x>0

Logarithmic model : f(x) (30 - 1/2ln (26/25) Ln(x)

domain: no idea.

looking forward to your repsonse!

 

[MarkFL]

Re: Are my answer correct?

For the quadratic model, you want:

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

where:

\(\displaystyle f(0)=a(0)^2+b=b=20\)

\(\displaystyle f(1)=a(1)^2+b=a+b=20.5\)

I would solve by substitution.

For the logarithmic model, we will have to pick a value with which to shift the graph to the left since $\ln(0)$ is undefined.

I would choose:

\(\displaystyle f(x)=a+b\ln(x+1)\)

where:

\(\displaystyle f(0)=a+b\ln(0+1)=a=30\)

\(\displaystyle f(1)=a+b\ln(1+1)=a+b\ln(2)=30.5\)

I would use substitution here as well.

 

[needalgebra]

Answers:

1) f(x) = 0.5x^2 + 20
2) f(x)=30+0.72[ln(x+1)1) f(x)=ax2+b
f(0)=a(0)2+b=b=20
f(1)=a(1)2+b=a+b=20.5 => a = 20.5 - 20 = 0.5 so,
f(x)=ax2+b f(x) = 0.5x^2 + 20

2) f(x)=a+bln(x+1)
f(0)=a+bln(0+1)=a=30
f(1)=a+bln(1+1)=a+bln(2)=30.5 =>b = (30.5-30) / lin(2) = 0.7213475204, so,
f(x)=a+bln(x+1) f(x)=30+0.72[ln(x+1)]

 

[MarkFL]

Looks good, although I would choose to express the parameter $b$ for the logarithmic model in exact form:

\(\displaystyle f(x)=30+\frac{\ln(x+1)}{2\ln(2)}\)

This would allow you to use the change of base formula to write:

\(\displaystyle f(x)=30+\log_4(x+1)\)

You probably want to stick with the first form though for using the computer to generate a graph.

 

[needalgebra]

the graph was going to be my next question! - sorry for being a pain in the butt. :p

How would would i put my answers into WA?