Recent content by 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?

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 !

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?

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??

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...

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...