We have an important chemsitry test due and I want to make sure that I do not mess up on my algebra. I am dealing with the integrated rate law, but my question is purely algebraically in nature. I hope, somebody can help me.

Question

1) ln(A) = -kt+ln(B)

solve for t
solve for k
solve for B

2) 1/(B) = kt + 1/(B)

solve for k
solve for t
solve for B

Thank you so much for your help. It is greatly appreciated

Answer 1

1)
ln(A) = -kt+ln(B)
kt = lnB - lnA
kt = ln(B/A)
t = ln(B/A) / k

k = ln(B/A) / t

lnB = lnA + kt
B = e^(lnA + kt)

2)
1/B = kt + 1/B
0 = kt
t = 0
or
k = 0
B is any value except B≠0
(are you sure you have no typo in 2) ?

Answer 2

2) should be 1/(B) = kt+1(A)
I am dealing with concentrations and A and B are just placeholders for the concentrations.
Thanks

Answer 3

2) I thought there was something wrong with your first typing of the question

1/B = kt + 1/A
multiply each term by AB, the LCD
A = ABkt + B
A - B = ABkt
t = (A-B)/(ABk)
k = (A-b)/(ABt)

from: A = ABkt + B
A = B(Akt + 1)
B = A/(Akt + 1)

Answer 4

kt00 b =2

Answer 5

Solve:
3x(x + 4) + 3(x + 4) = 0
A) 0 and -4
B) -1 and 4
C) -1 and -4
D) 1 and 4

Answer 6

3/7x-4=-1