Logarithm functions

[zenith mantle]

How to form the functions for these graphs?

[boreal parrot]

What are the slopes of those "lines"?

And what do those slopes mean with those axes?

[zenith mantle]

It´s not given

I am supposed to calculate it. But I dont know how

The left graph has the points A(4/3^-5) B(1/3^1)
The right graph has the points C(2^9/2^20) D(2^3/2^8)

That´s the task. 😦

[boreal parrot]

I mean their slopes on this graph

You know, rise/run

Count the units up and divide by the units right, between the two points

[zenith mantle]

you mean (y2-y1)/(x2-x1) Omg i know what I did wrong..... I haven´t counted correctly 🤦‍♀️

@boreal parrot

ok no. still not right i think

@boreal parrot

[novel epoch]

notice the y scale is a log scale,

[zenith mantle]

exactly

[novel epoch]

every time you go 3 units over you divide by 729

so every time you go 1 unit over you divide by 9

[zenith mantle]

but i dont know how to deal with it

Ok wait let me think

[novel epoch]

you could say you multiply by 1/9

[zenith mantle]

but why

[novel epoch]

well

because going from 3^1 to 3^-5 is dividing by 729

a straight line on a lin-log graph is an exponential function

[zenith mantle]

how do you know T^T

[novel epoch]

what is T^T

i just know 3^6

[zenith mantle]

Its a crying face

[zenith mantle]

No clue how you get 3^6 😭

[novel epoch]

because

3^1/3^-5 = 3^6

took me a moment

[zenith mantle]

ohh, so I always have to divide y2/y1 and x2/x1 if its a log scale

but how do i calculate the slope then? @novel epoch

[novel epoch]

no

y2/y1 and x2 - x1

[zenith mantle]

If the x-axis is also logarithmically scaled, then you would calculate the slope the way I said, right?

[novel epoch]

yes

a straight line on a log-log is a polynomial whose degree is the slope

i believe

[zenith mantle]

but how do i calculate the slope now 😅 Its y2-y1/x2-x1

its really challenging

[novel epoch]

i'm not 100% sure what you mean by calculate the slope

[novel epoch]

this is the slope

[zenith mantle]

I need to establish the function.

[novel epoch]

every time you go 1 unit over you multiply by 1/9

you know how to write function every time you go over 1 you multiply by 2 i assume?

[zenith mantle]

f(x)=2x?

[novel epoch]

this is constant*2^x

thats the definition of 2^x

[zenith mantle]

Ok, so as i said 2x?

[novel epoch]

2x =/= 2^x

[zenith mantle]

ahhh opssie forgot

meant 2^x

[novel epoch]

so now you can start writing the function here

[zenith mantle]

the solution is supposed to be 16x^-3. Dont get that

[novel epoch]

uhhh what

[zenith mantle]

yes 😭

[novel epoch]

that's the second one i assume

i am talking about the first one

[zenith mantle]

oh trueeee

[novel epoch]

except that's not even the second one

idk what you're talking about

[zenith mantle]

my bad i mixed thing up 😦

This whole topic is confusing me

[zenith mantle]

so it must be something 3^x ?

[novel epoch]

something like that

but not that

[zenith mantle]

yeah its not the final function 😦

Dont know how to do that

[novel epoch]

going over 1 means multiplying by 1/9

so you have 1/9^x

except that's not the final answer

as you can confirm by plugging in some points

the trick is that that has what you can think of as the correct "slope", but the wrong "intercept"

it has the right rate of decrease but does not start at the correct value

[zenith mantle]

yes, so far i can follow

[novel epoch]

so figure out how to modify 1/9^x so it fits the points on the given line

[zenith mantle]

f(x)= c1/9^x then plugging in some points. Point(1/3^1)
3^1= c1/9^1

c=27

f(x)= 27*1/9^x

@novel epoch

??

[novel epoch]

yes

27/9^x

also written as 3^(3-2x)

do you need help with the other one you posted too

[zenith mantle]

27/9^x not 27*1/9^x?

[zenith mantle]

probably XD

[novel epoch]

$\frac{27}{9^x} = 27\cdot \frac{1}{9^x}$

[warm trout]

cute rizzly bear (won't eat you)

[novel epoch]

same thing

[zenith mantle]

okay puhhh

[novel epoch]

alright
multiplying the x value by ..? multiplies the y value by ..?

[zenith mantle]

for y axis: 2^20/2^8=4096
and

x axis: 2^9/2^3=64

[novel epoch]

note that 4096 = 64^2

multiplying the x by c multiplies the y by c^2

[zenith mantle]

my brain is overwhelmed ummm.

[novel epoch]

ok lets do it in a different way

[zenith mantle]

is it the same function like y=c*a^x?

[novel epoch]

notice that if the powers of 2 on the x and y axes were not in the exponent

it would be linear function

you can take log2(y) and log2(x) to extract those powers

and find how they are related

[zenith mantle]

okay

log2(20)=4,32
log2(9)=3,17
A(3,17/4,32)
log2(3)=1,58
log2(8)=3
B(1,58/3)
f(x)= ax+b

[novel epoch]

no

it's not log2(20)

[zenith mantle]

sadly

[novel epoch]

it's log2(2^20)

[zenith mantle]

ahh dammit

log2(2^20)=20
log2(2^9)=9
A(9/20)
log2(2^3)=3
log2(2^8)=8
B(3/8)
f(x)= ax+b

[novel epoch]

(9,20) and (3,8)

[zenith mantle]

y2-y1/x2-x1= (20-8)/9-3=2

f(x)= 2x+b plugging in the points again

20=2*9+b

b=2

f(x)=2x+2

[novel epoch]

yes

log2(y) = 2log2(x) + 2

do you get why

[zenith mantle]

F*ckkkk yeaaaaah

[novel epoch]

alright now you can. solve

[zenith mantle]

thats kinda confusing me but i am happy i got so far with your help

[novel epoch]

you found that log2(y) and log2(x) are linearly related

[zenith mantle]

yyes

[zenith mantle]

I dont understand 2log2(x)+2

the two before the log

[novel epoch]

you found f = 2x + 2

substituting log2(y) for f and log2(x) for x

[zenith mantle]

ohhh now i see

thanks a loooooot

[novel epoch]

did you figure it out

[zenith mantle]

i did. I am so happy thaankkk you !!!

[novel epoch]

yw

[zenith mantle]

@novel epoch this is my last question. How do you got 1/9 before?

[novel epoch]

going over 3 multiplies by 1/729

= 1/9*1/9*1/9

if going over 3 multiplies by 1/(a^3) it is reasonable to suggest that going over 1 multiplies by 1/a

[zenith mantle]

root3(1/729)=1/9

Alright i got it

[zenith mantle]

@novel epoch 😢 Could you help me again

the log scaled graph again

I dont even find anything on the internet that explains this

So far I got: (2^7-2^-5)/(2^-1/2^3)= -17,0625=-273/16 for the slope but thats obviously wrong

[novel epoch]

as before, on a log-log graph you want to find the relation between log(y) and log(x)

aka the relationship between the exponents of the y scale and the exponents of the x scale

where you do log base (whatever the base is)

here it appears to be (-1,7) and (3,-5)

[zenith mantle]

it seems a bit different than the other example we did yesterday. The last one was linear and this one is not as the results show and that confuses me 😭

[novel epoch]

we did 2 yesterday

a log-lin plot and a log-log plot

[zenith mantle]

The other one yesterday was only scaled on the y axis.

[zenith mantle]

exactly and I amfocusing on the log log plot

coz its the same in thise task

[novel epoch]

when it's log, you don't subtract, you divide on that axis

that may be what is confusing you

except that's just not a very helpful way to figure it out

[zenith mantle]

I tried that as well but it seemed to be wrong

[novel epoch]

do this

[zenith mantle]

So,
y-axis: 2^7/2^-5= 4096
x-axis: 2^3/2^-1=16

y-axis: 1/4096 root2(1/4096)= 1/64
x-axis: 1/16 root2(1/16) = 1/4

@novel epoch

[novel epoch]

.

[zenith mantle]

but what do i do with that

i dont know how to proceed

[novel epoch]

relate log2(y) with log2(x)

[zenith mantle]

wh is that so complicating 😭

[novel epoch]

maybe here is a less problem solvy more memorizy way

a straight line on a log log plot is as i already said, a polynomial

ax^b

[zenith mantle]

Yes and then I have the point as you mentioned above:

[novel epoch]

you can solve for a and b given the 2 points

[zenith mantle]

f(x)=ax^b

I. log2(2^7)= alog2(2^-1)^b
II. log2(2^3)= alog2(2^-5)^b

[novel epoch]

no

y = ax^b

or log2(y) = blog2(x) + m

those are equivalent

you can solve with either

[zenith mantle]

but yesterday i was supposed to keep the base in :

[novel epoch]

that is when you are solving this way

[zenith mantle]

I am hella confused

[novel epoch]

ok, that's my fault, just solve y = ax^b

[zenith mantle]

I. 7 = a* (-1)^x
II. -5=a*3^x

@novel epoch

[novel epoch]

[zenith mantle]

thats how i am feeling

hahhaa

[novel epoch]

y is not 7

y is 2^7

etc

and it's not to the power of x

it's x to the power of b

we have points (2^-1, 2^7) and (2^3, 2^-5)

2^7 = a*(2^-1)^b
2^-5 = a*(2^3)^b

it seems this is a bit annoying to solve

[zenith mantle]

it is 😭

[novel epoch]

take the ratio between the 2 equations

first equation divided by second equation

2^12 = (2^-1)^b/(2^3)^b

exponent rules

[zenith mantle]

2^12=1/16^b
=> 2^12=(2^-4)^b |log2(2^b)
=> 12=-4b | :(-4)
=> -3 =b

[novel epoch]

yes

[zenith mantle]

f(x)= ax^3
2^7=a*(2^-1)^-3 | :*(2^-1)^-3
a=16

y= 16x^-3

we did it

omg

can i always do it like that or are there exceptions XD

[novel epoch]

if it's a straight line on a log-log plot, that should work

[zenith mantle]

I probably annoyed you already. 😅 Thanks for everything