Question

 log6 (x+1)− log6 x =log6 29 

Answer 1

$log_6(x+1)-log_6x=log_629;\ D:x+1 > 0\ \wedge\ x > 0\Rightarrow x\in\mathbb{R^+}\\\\log_6\frac{x+1}{x}=log_629\iff\frac{x+1}{x}=29\\\\29x=x+1\\\\29x-x=1\\\\28x=1\ \ \ \ /:28\\\\x=\frac{1}{28}\in D\leftarrow solution$

Answer 2

$D:x+1>0 \wedge x>0\\ D:x>-1 \wedge x>0\\ D:x>0\\ \log_6(x+1)-\log_6 x=\log_6 29\\ \log_6(x+1)=\log_6 29+\log_6x\\ \log_6(x+1)=\log_6 29x\\ x+1=29x\\ 28x=1\\ x=\frac{1}{28}$