how do you solve ln(x+1)-ln(x)=ln4

Question

ln(x+1)-ln(x)=ln4

Answer 1

ln(x+1) -ln(x) = ln(4)

 

if you remember the basic rules of logs

 

1) log_b(mn) = log_b(m) + log_b(n)

2) log_b(m/n) = log_b(m) – log_b(n)

3) log_b(mⁿ) = n · log_b(m)

 

So if we look at your eqn, we can use rule 2 

 

ln(x+1) - ln(x) = ln((x+1)/x)) = ln 4

 

Then if you remember that to solve an eqn for a given variable you need to undo whatever has been done to the variable. In our case, utilizing the inverse of the ln(x)

 

f (f ^-1(x)) = e^ln(x) = x

 

we can apply that to both sides of the eqn

 

e^ln((x+1)/x)) = e^{ln 4}  ⇒  (x+1)/x = 4

 

then solving this for x is easier

 

(x+1)/x) = 4

(x+1)/x •  x = 4 • x

x+1 = 4x

x+1 -x = 4x -x

1 = 3x

1/3 = 3x/3

x = 1/3

 

Hope this helps

Answer 2

ln[(x+1)/x] = ln4

So, (x+1)/x = 4

x+1 = 4x

Answer: x = 1/3