Question 255524
{{{log(x) = log((x+1)) + 1}}}

The problem here is that we have two logarithms.  Our first step is too subtract log(x+1) from both sides.

{{{log(x) - log((x+1)) = 1}}}

Now, using one of the laws of logarithms (log(a) - log(b) = log(a/b)) we have:

{{{log(x/(x+1)) = 1}}}

Now we rewrite the problem in exponential form as:

{{{10^1 = x/(x+1)}}}

Now multiply both sides by x+1 :

{{{10(x+1) = x}}}

{{{10x+10 = x}}}

{{{9x + 10 = 0}}}

{{{9x = -10}}}

{{{x = -10/9}}}

However, you cannot take the log of a negative number.  Thus, this equation has no solution.