I need to find the function of f with given data

[mimi.janson]

Homework Statement

Hi

I need to determine ragulatory for f, where the expiration time is measured as the function of the temperature.

The function type is f(t)=b*a^t

i get two pair of datas which are that for -15 °C the expiration day will be after 200 and by -5 ° the expiration day will be after 10 days

Homework Equations

I know that t is the temperature and f(t) must be the days, and i need to determine a regulatory for f, which means i need to find a and b i think

so i should maybe put my given informations in f(t)=b*a^t which will be

200=b*a^-15

and

10=b*a^-5

then i shall isolate b

200/a^-15 =b and 10/a^-5 =b

these to get put up agains each other so a can be solved.

200*a^-5 =10*a^-15

now i solve a

200=10*a^-20

a=tenth√(200/10)

a=1.16

But it is wrong and i don't get why so can someone please help me??

The Attempt at a Solution

 

[gneill]

Homework Statement

Hi

I need to determine ragulatory for f, where the expiration time is measured as the function of the temperature.

The function type is f(t)=b*a^t

i get two pair of datas which are that for -15 °C the expiration day will be after 200 and by -5 ° the expiration day will be after 10 days

Homework Equations

I know that t is the temperature and f(t) must be the days, and i need to determine a regulatory for f, which means i need to find a and b i think

so i should maybe put my given informations in f(t)=b*a^t which will be

200=b*a^-15

and

10=b*a^-5

then i shall isolate b

200/a^-15 =b and 10/a^-5 =b

these to get put up agains each other so a can be solved.

200*a^-5 =10*a^-15

now i solve a

200=10*a^-20

Oops! Check your exponent work; you want to multiply both sides by a⁵.

Otherwise, you seem to be doing okay

 

[mimi.janson]

Oops! Check your exponent work; you want to multiply both sides by a⁵.

Otherwise, you seem to be doing okay

thank youu but sorry i cannot see on which step of the ones i have to do it but is it right if i think what you are telling me is what happens in the third last of my steps?

200*a^-5*a⁵ =10*a^-15*a⁵?? but i don't know how to solve it when it is written like this

 

[gneill]

thank youu but sorry i cannot see on which step of the ones i have to do it but is it right if i think what you are telling me is what happens in the third last of my steps?

200*a^-5*a⁵ =10*a^-15*a⁵?? but i don't know how to solve it when it is written like this

Yes, that's where I see a problem. You wanted to get rid of the a^-5 on the left so you divided both sides by a^-5, which is the same thing as multiplying both sides by a⁵. But you made a mistake handling the exponents on the right hand side.

What's $a^{-15}\cdot a^5$ ?

 

[mimi.janson]

Yes, that's where I see a problem. You wanted to get rid of the a^-5 on the left so you divided both sides by a^-5, which is the same thing as multiplying both sides by a⁵. But you made a mistake handling the exponents on the right hand side.

What's $a^{-15}\cdot a^5$ ?

that would be a^-75?

so then i solve a

200=10*a^-75

a=-75th √(200/10)

but is there not a rule saying that i cannot take the -xsquaroot of something and i have -75?



and should i also keep going and find b like or does one thing need to be undefined to have a function?

 

[gneill]

Okay, you should review your laws of exponents:

$a^x \cdot a^y = a^{x+y}$

$\frac{a^x}{a^y} = a^{x-y}$

 

[mimi.janson]

Okay, you should review your laws of exponents:

$a^x \cdot a^y = a^{x+y}$

$\frac{a^x}{a^y} = a^{x-y}$

then i say -15 + 5 which would be -10? but you know its the - infront of 10 that's bugging me...because it can't be a minus since i have to take the square root of it later

 

[gneill]

then i say -15 + 5 which would be -10? but you know its the - infront of 10 that's bugging me...because it can't be a minus since i have to take the square root of it later

No, you shouldn't have to take the square root of a negative number. From your previous work, and repairing the exponent math, the equation is:

$200 = 10 a^{-10}$

$20 = a^{-10}$

You can proceed using exponent rules, or use logarithms.

 

[mimi.janson]

No, you shouldn't have to take the square root of a negative number. From your previous work, and repairing the exponent math, the equation is:

$200 = 10 a^{-10}$

$20 = a^{-10}$

You can proceed using exponent rules, or use logarithms.

well I am not sure about how to use the logarithm, but i tried solving it with a program that we also use, and what do i do if i get 5 relsults put in a parensesis and under each other

0
0
1,075
-0,538+0,93li (i don't know what li strands for)
-0,538-0,93li

i believe i should use that a is 1,075 so do i need to find the b too? because i don't get how to make an f function out of it now

 

[gneill]

well I am not sure about how to use the logarithm, but i tried solving it with a program that we also use, and what do i do if i get 5 relsults put in a parensesis and under each other

0
0
1,075
-0,538+0,93li (i don't know what li strands for)
-0,538-0,93li

i believe i should use that a is 1,075 so do i need to find the b too? because i don't get how to make an f function out of it now

The "i" stands for $\sqrt{-1}$. Those are called "imaginary" numbers (as opposed to the "real" numbers that you're used to), and you'll learn about them at a later date.

What you're looking for here will be real numbers, so when you take the nth root of a real number you want what's called the "principle value".

Here you can continue to simplify your expression to make it simpler to solve.

$20 = a^{-10}$

$20^{-\frac{1}{10}} = a~~~~~~$ {raised both sides to the -1/10 to isolate a}

$a = \frac{1}{20^{1/10}}~~~~~~$ {moved "20" to denominator to clear negative}

So you need to find the 10th root of 20, then take its inverse.

Once you have a value for a you can plug it into either of the original expressions to solve for b.

 

[mimi.janson]

The "i" stands for $\sqrt{-1}$.

$20 = a^{-10}$

$20^{-\frac{1}{10}} = a~~~~~~$ {raised both sides to the -1/10 to isolate a}

$a = \frac{1}{20^{1/10}}~~~~~~$ {moved "20" to denominator to clear negative}

So you need to find the 10th root of 20, then take its inverse.

Once you have a value for a you can plug it into either of the original expressions to solve for b.

i don't get how you go from a^-10 to -1/10 ? how do you get the 1 ?


ok so i got the 10th root of 20 to be 1,349


and i put it in the expression for b so


200/1,349^-15=1,783*10⁴

and

10/1,349^-5=44,675

but this shows i did something wrong since my two attempts at finding b were not the same or the two pairs of numbers I've got

 

[gneill]

i don't get how you go from a^-10 to -1/10 ? how do you get the 1 ?

Exponent rules (you should become very familiar with them):
$$(a^x)^y = a^{x\cdot y}$$
Here the exponent of a is -10 and you'd like it to be 1. So $(-10) \cdot \left(-\frac{1}{10}\right) = 1$. The same operation is applied to the 20 on the other side of the equation in order to keep the equation balanced.

ok so i got the 10th root of 20 to be 1,349

Hmm, that seems unlikely. Surely 20th root of 20 should be less than 20 itself?
How did you obtain the value?

 

[mimi.janson]

Exponent rules (you should become very familiar with them):
$$(a^x)^y = a^{x\cdot y}$$
Here the exponent of a is -10 and you'd like it to be 1.

I said the 10th root of 200/10...but i know its wrong now...i just tried again and got a = 0,741 isn't that right and b=2,3?

I know i might be irritating because i keep asking and i really don't hope I am bothering you too much but i really still don't get the exponent rule you used last, they are even shown in my math book but i simply don't get it.

yes i would like it to be one but how did you know that i would have to say - 1/10 ? i mean i wouldn't know that you had to use a 1, and when i compare what is done it does not look like that rule at all a^x*y i mean it even contains a division where we are using multiplikation

 

[gneill]

I said the 10th root of 200/10...but i know its wrong now...i just tried again and got a = 0,741 isn't that right and b=2,3?

Yes, that looks good.

I know i might be irritating because i keep asking and i really don't hope I am bothering you too much but i really still don't get the exponent rule you used last, they are even shown in my math book but i simply don't get it.

yes i would like it to be one but how did you know that i would have to say - 1/10 ? i mean i wouldn't know that you had to use a 1, and when i compare what is done it does not look like that rule at all a^x*y i mean it even contains a division where we are using multiplikation

Okay, let's consider the object $a^{-10}$. It's exponent is currently -10. If I wish to make the exponent 1, I need to divide it by -10, right? But dividing by -10 is the same as multiplying by $\frac{1}{-10}$. Putting it in the form of the given exponent rule, letting $x = -10$ and $y = -\frac{1}{10}$, then

$$(a^x)^y = a^{xy} = a^{-10 \frac{1}{-10}} = a^\frac{-10}{-10} = a^1 = a$$

So I chose -1/10 in order to clear the -10 from the exponent of a.

And of course, if you do something to one side of an equation, you have to do the same thing to the other side in order to retain equality.

 

[mimi.janson]

Yes, that looks good.

Okay, let's consider the object $a^{-10}$. It's exponent is currently -10. If I wish to make the exponent 1, I need to divide it by -10, right? But dividing by -10 is the same as multiplying by $\frac{1}{-10}$. Putting it in the form of the given exponent rule, letting $x = -10$ and $y = -\frac{1}{10}$, then

$$(a^x)^y = a^{xy} = a^{-10 \frac{1}{-10}} = a^\frac{-10}{-10} = a^1 = a$$

So I chose -1/10 in order to clear the -10 from the exponent of a.

And of course, if you do something to one side of an equation, you have to do the same thing to the other side in order to retain equality.

so if i divided -10 by -10 it would also be the right and same result?

 

[gneill]

so if i divided -10 by -10 it would also be the right and same result?

Sure. Note that on the other side of the equation you have 20, with an assumed exponent of 1. Dividing the exponent by -10 gives $20^{-1/10}$, the same as before.