f(5)=12 & f(10)=18: f(20)=(?) for Exponential, Power and linear functions

C

chuckB12

New member
Joined
Mar 24, 2021
Messages
3
Learning online on my own is proofing difficult... appreciate any help... cheers...:D
If a function f has values f(5)=12 and f(10)=18 , find f(20)=(?) if,
a) Exponential Function;
b) Linear Function;
c) Power Function.
 
chuckB12 said:
Learning online on my own is proofing difficult... appreciate any help... cheers...:D
If a function f has values f(5)=12 and f(10)=18 , find f(20)=(?) if,
a) Exponential Function;
b) Linear Function;
c) Power Function.
Click to expand...
We'll need to know what help you need. If you need a place to start, that would be to find the definitions you have been given for each type of function, which may or may not be what I would guess.

Once you've told us the three forms (for instance, the linear function is probably something like y = ax + b), then in each case you will be replacing x and y with the two given pairs, resulting in a pair of equations you can solve for the parameters (e.g. a and b).

Please show whatever you can do and what you have been taught, and we can discuss the details.
 
You need to read

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...
www.freemathhelp.com

What is the standard form of an exponential function?

What is the standard form of a linear function?

What is the standard form of a power function?
 
The linear and exponential I have solved, my problem now is with the power case, I applied

f(x)=axk

but I might be doing something wrong when trying to sort it... tks for your help guys...:D
 
Dr.Peterson said:
We'll need to know what help you need. If you need a place to start, that would be to find the definitions you have been given for each type of function, which may or may not be what I would guess.

Once you've told us the three forms (for instance, the linear function is probably something like y = ax + b), then in each case you will be replacing x and y with the two given pairs, resulting in a pair of equations you can solve for the parameters (e.g. a and b).

Please show whatever you can do and what you have been taught, and we can discuss the details.
Click to expand...


I have solved the Exponential and the linear, now trying the power function, applying the formula f(x)=axk, but I am having some trouble in solving it... thank you Dr. Peterson

That is what I have for the power function so far... but it's wrong somewhere... tks...

Power Function: y = Ax^k
f(5) = 12
12 = A(5)^k

f(10) = 18
18 = A(10)^k

Solve 2 equations, 2 unknowns:
12 / 5^k = 18 / 10^k
12 / 18 = 10^k / 5^k
2/3 = 2^k•5^k / 5^k
2/3 = 2^k

k = log(base 2) (2/3)
k = ln(2/3) / ln(2)
k = (ln2 - ln3) / ln2
k = 1 - ln3/ln2
k = -0.585

A = 12 / 5^(-0.585)
A = 30.8

y = 30.8x^(-0.585)
y(20) = 30.8(20)^(-0.585)
y(20) = 5.34
 
Last edited:
chuckB12 said:
I have solved the Exponential and the linear, now trying the power function, applying the formula f(x)=axk, but I am having some trouble in solving it... thank you Dr. Peterson

That is what I have for the power function so far... but it's wrong somewhere... tks...

Power Function: y = Ax^k
f(5) = 12
12 = A(5)^k

f(10) = 18
18 = A(10)^k

Solve 2 equations, 2 unknowns:
12 / 5^k = 18 / 10^k
12 / 18 = 10^k / 5^k
2/3 = 2^k•5^k / 5^k
2/3 = 2^k

k = log(base 2) (2/3)
k = ln(2/3) / ln(2)
k = (ln2 - ln3) / ln2
k = 1 - ln3/ln2
k = -0.585

A = 12 / 5^(-0.585)
A = 30.8

y = 30.8x^(-0.585)
y(20) = 30.8(20)^(-0.585)
y(20) = 5.34
Click to expand...
The work looks good without checking every detail; let's check it out:

y = 30.8x^(-0.585)

f(5)=30.8*5^(-0.585)=12.013049 which is close to 12
f(10)=30.8*20^(-0.585)=5.33885553 which is not close to 18

So something went wrong with the determination of k. And when I look more closely at your work there, I see a silly mistake in the first couple lines. See if you can find it!

Also, you might observe that the function should be an increasing function, so something about k should have felt wrong.
 
Here is what I mean. If you simplify to logs immediately, the algebra is somewhat less messy and so less error prone. The base of course is irrelevant.

[MATH]f(5) = 12 \text { and } f(10) = 18 \text { and } y = f(x) = ax^k.[/MATH]
[MATH]\therefore ln(y) = ln(a) + k \ ln(x) \implies ln(12) = ln(a) + k \ ln(5) \text { and } ln(18) = ln(a) + k \ ln(10) \implies[/MATH]
[MATH]ln(12) - ln(18) = k\{ln(5) - ln(10)\} \implies k = \dfrac{ln(12) - ln(18)}{ln(5) - ln(10)} =[/MATH]
[MATH]\dfrac{ln(12)- ln(18)}{ln(5/10)} = \dfrac{ln(12) - ln(18)}{ln(1/2)} = \dfrac{ln(12) - ln(18)}{- ln(2)} = \dfrac{ln(18) - ln(12)}{ln(2)} = \dfrac{ln(18/12)}{ln(2)} = \dfrac{ln(1.5)}{ln(2)}.[/MATH]
[MATH]\therefore ln(12) = ln(a) + k \ ln(5) \implies ln \ (a) = ln(12) - ln(5^k) = ln \left ( \dfrac{12}{5^k} \right ) \implies a = \dfrac{12}{5^k}.[/MATH]
[MATH]\therefore f(x) = ax^k, \text { where } a = \dfrac{5}{12^k} \text { and } k = \dfrac{ln(2/3)}{ln(1/2)}.[/MATH]
Now that is still a bit messy so we should check.

[MATH]f(5) = a * 5^k = \dfrac{12}{5^k} * 5^k = 12. \ \checkmark.[/MATH]
[MATH]f(10) = a * 10^k = \dfrac{12}{5^k} * 5^k * 2^k = 12 * 2^k.[/MATH]
But what is [MATH]2^k[/MATH]?

[MATH]z = 2^k \implies ln(z) = k \ ln(2) = \dfrac{ln(2/3) * ln(2)}{ln(1/2)} = \dfrac{\{ ln(2) - ln(3)\} * ln(2)}{- ln(2)} = ln (3) - ln(2) = ln(1.5) \implies[/MATH]
[MATH]z = 1.5 \implies 2^k = 1.5 \implies 12 * 2^k = 18 \implies f(10) = 18. \ \checkmark.[/MATH]
It is not conceptually hard, but it is fussy algebra.
 
You must log in or register to reply here.
Top