_{b}(mn) = log

_{b}(m) + log

_{b}(n)

2) log

_{b}(m/n) = log

_{b}(m) – log

_{b}(n)

3) log

_{b}(m

^{n}) = n · log

_{b}(m)

^{ -1}(x)) = e

^{ln(x)}= x

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

^{ln 4}⇒ (x+1)/x = 4

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

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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}(m^{n}) = n · log_{b}(m)

2) log

3) log

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

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

So, (x+1)/x = 4

x+1 = 4x

Answer: x = 1/3

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