Question

Use the first and last points to find algebraically the particular equation of the natural logarithmic function that fits the point

Answer 1

Table: x= 3.6, 14.4, 57.6, 230.4, 921.6
y= 1, 2, 3, 4, 5

Answer 2

\[y=\log_{4} \frac{ 10 }{ 9 }x\]

Answer 3

That is the answer
But how do I get the answer? Thanks!

Answer 4

did you try to enter the last and first points into \[y=\log_a(bx)\]
and solve for a and b?

Answer 5

But how do I do that. My teacher didn't teach logarithms last year in Algebra 2, that's why this concept is new to me

Answer 6

you can plug in the points
\[(3.6,1) \text{ and } (921.6,5)\]
since it said plug and the first and last points

Answer 7

logarithmic is the inverse of a exponent.

Answer 8

(x,y) replace x in my equation with the first number of the pair
and replace y in my equation with the second number of the pair

you will have two equations

Answer 9

@myininaya you will have to tell him what a logarithm is because he said he never had it in a previous course.

Answer 10

I understand what logarithm is, but I never performed any operations on them, that's why this is confusing for me

Answer 11

Will I need to create two equations for this problem?

Answer 12

\[1=\log_{a}(3.6x)\]

Answer 13

\[5=\log_{a}(921.6b) \]

Answer 14

like this? except the first equation: x should be b

Answer 15

Okay. So you have two equations. How would you simplify this? Any ideas?


Use this relation

\[ a^y=x\]

\[y = \log_{a} x\]

Answer 16

I think by dividing the a's

Answer 17

You log equations should be setup like

\[1 = \log_{a} (3.6x)\]
\[5= \log_{a} (921.6x)\]

Answer 18

shouldn't it be 3.6b instead of 3.6x

Answer 19

Yes you right my bad. Since we don't know b.

Answer 20

Let's get of the logs

\[1 = \log_{a} (3.6b)\]
\[5 = \log_{a} (921.6b)\]

Answer 21

I understand the setup now, but what do I do next?

Answer 22

solve for a?

Answer 23

Yes.
You can solve for either a or b.

Answer 24

Will logs cancel out? so it will be: a^1=3.6b and a^5= 921.6b

Answer 25

That is correct. Since you raise base a the logs are now gone. Now you need to solve for one of these variables

Answer 26

so it will be: a^-4 = (3.6/921.6)b

Answer 27

or do I need to flip the two equations

Answer 28

That will work but b should go away since your dividing two equations.

Answer 29

a^-4 = 3.6/921.6

Answer 30

And how to I get rid of the negative exponent?

Answer 31

Same as you would do with a non negative exponent. In this case you will take the negative root.

Answer 32

I get it, I just need to put 1/a^4

Answer 33

You can do that or
\[a=\sqrt[-4]{3.6/921.6} \]

Answer 34

a^4 = 256 a=4 Thanks!

Answer 35

Now we can find b.

Answer 36

and after I plug in 4 for B in one of the equations.\[1=\log_{a} (3.6b)\]

Answer 37

\[1=\log_{a}(14.4) \]

Answer 38

then a = log 14.4

Answer 39

a=1.1584

Answer 40

Actually you can use your previous result to get b. Early you got
\[a^1 = 3.6b\]

Since you know a you can plug a in here and solve for b.

Answer 41

4=3.6b b= 1.1

Answer 42

Good. You got b now.

Answer 43

to get rid of the decimal, I can multiply the fraction's numerator and denominator by 10

Answer 44

and then reduce it to 10/9

Answer 45

That is correct. Good job.

Answer 46

\[y=\log_{4}\left(\begin{matrix}10 \\ 9\end{matrix}\right)x\]

Answer 47

(10/9x)