Trouble with Log Laws and exponential functions

[pdas]

Please help i have these two questions and i am Stuck!
Question 1
The linear equation you have found (y=.771x+1.609) is in the form of y= mx +c. it should be however more appropreiatly be considered as being in the form, lny = mx + lnA. by using appropriat logarithmic and exponential laws rearange your equation (y=.771x+1.609) to make y the subject

I have no idea where to start with this Question.


Question 2
Bones A and B are x and y thousand years old respectivly. a Geiger Counter Confirms that bone A contains 3 Times as much C-14 as Bone be What can be Said about the ages of Bones A and B. The exponential equation that is given for the c-14 decay rate is a=15.3 x 0.886^t

i Have a feeling Question 2 has something to do with similtanious equations and i have gathered this so far.
3a = 15.3x0.883^x......A
a = 15.3x0.886^y......B

thanks you any help is very much appreciated

 

[tiny-tim]

welcome to pf!

hi pdas! welcome to pf!

(try using the X² icon just above the Reply box )

… rearange your equation (y=.771x+1.609) to make y the subject

i don't understand … y is the subject of y = 0.771x + 1.609

i Have a feeling Question 2 has something to do with similtanious equations and i have gathered this so far.
3a = 15.3x0.883^x......A
a = 15.3x0.886^y......B

That's right …

and when you have two simultaneous equations, combine them into one equation, with y on the left and x on the right

 

[pdas]

i think it means make y the subject in the lny = mx + lnA equation ie get rid of the natural logs but i am not sure on how to do it??

is there any chance that you could help me a bit more with question two I am not sure how to continue i have made equation A into
log(a/15.3)/log0.886 = y is this right??

 

[pdas]

I was looking at it a tried to do it this way is this right??
a = 15.3x0.886^(log(a/15.3)/log0.886) but then how do i continue to solve??

 

[tiny-tim]

hi pdas!

(just got up :zzz: …)

i think it means make y the subject in the lny = mx + lnA equation

oh i see!

ok, do e-to-the on both sides (so the LHS becomes e^lny, which is y)

3a = 15.3x0.883^x......A
a = 15.3x0.886^y......B

you need to eliminate "a" from those two equations …

so divide one by the other! ​

 

[pdas]

ok Question 1 makes sense now.

i think my brain just clicked :)
So if i divide the 2 equations i get
3=x-y
then make it x=3+y
then divide it the other way around so u get
1/3= y-x
1/3=y-(3-y)

but then where do i go?

 

[tiny-tim]

So if i divide the 2 equations i get
3=x-y

no, you don't!

write it out carefully, then divide

 

[pdas]

i remember now keep the base take the indices
so is it 3=.886^(x-y)?

then log3/log.886=(x-y)?


and after an hour of looking at question 1 again i am lost i get the e^(lny)=mx + but i don't see how lnA is related to c?? is c = ln A??

thank you so much for helping me !

 

[tiny-tim]

i remember now keep the base take the indices
so is it 3=.886^(x-y)?

then log3/log.886=(x-y)?

Woohoo! ​

and after an hour of looking at question 1 again i am lost i get the e^(lny)=mx + but i don't see how lnA is related to c?? is c = ln A??

you must write everything out fully!

not because the examiner needs to see it, but because you need to see it!

ok, in full:

lny = mx + lnA,

so e^lny = e^{mx + lnA}

so e^lny = e^mxe^lnA

so … ? ​

 

[pdas]

so y = e^mx.e^lnA

then with Question 2 is it then true that you can only find the difference between the two values? not the values them selfs?

 

[tiny-tim]

hi pdas!

(just got up :zzz: …)

so y = e^mx.e^lnA

(please use the X² tag just above the Reply box )

that's still not finished …

simplify it some more ​

then with Question 2 is it then true that you can only find the difference between the two values? not the values them selfs?

("themselves" )

yup!