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.

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

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) ?

2) should be 1/(B) = kt+1(A)

I am dealing with concentrations and A and B are just placeholders for the concentrations.
Thanks

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)

kt00 b =2

Solve:

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

3/7x-4=-1

I'd be happy to help you with your algebraic questions related to the integrated rate law. Let's go through each question step by step:

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

To solve for t:
Step 1: Start by isolating -kt on one side of the equation by subtracting ln(B) from both sides:
ln(A) - ln(B) = -kt

Step 2: Divide both sides of the equation by -k to isolate t:
t = (ln(B) - ln(A)) / k

To solve for k:
Step 1: Start by isolating -kt on one side of the equation by subtracting ln(B) from both sides:
ln(A) - ln(B) = -kt

Step 2: Divide both sides of the equation by -t to isolate k:
k = (ln(A) - ln(B)) / -t

To solve for B:
Step 1: Start by bringing -kt to the other side of the equation:
ln(A) - ln(B) = -kt

Step 2: Take the antilog or exponential of both sides of the equation using the base e (natural logarithm base) to eliminate the ln:
e^(ln(A) - ln(B)) = e^(-kt)

Step 3: Simplify the equation:
A / B = e^(-kt)

Step 4: Rearrange the equation to isolate B:
B = A / e^(-kt)

2) 1/B = kt + 1/B

To solve for k:
Step 1: Subtract 1/B from both sides of the equation:
1/B - 1/B = kt

Step 2: Simplify the equation:
0 = kt

Step 3: Since anything multiplied by 0 is equal to 0, k can be any value. There is no unique solution for k in this case.

To solve for t:
Step 1: Subtract 1/B from both sides of the equation:
1/B - 1/B = kt

Step 2: Simplify the equation:
0 = kt

Step 3: Divide both sides of the equation by k:
0/k = t

Step 4: Since anything divided by 0 is undefined, there is no valid solution for t in this case.

To solve for B:
Step 1: Start by subtracting kt from both sides of the equation:
1/B - kt = 1/B

Step 2: Rearrange the equation to isolate B:
B = 1 / (1/B - kt)

I hope these explanations help you understand how to solve for t, k, and B in these algebraic expressions. Good luck with your chemistry test!