Question

This was my very last homework question and i'm stumped. log_4(x+1)-log_4(x)=3

Answer 1

$\log_4(x+1)-\log_4x=3 \qquad(x>0)\\ \log_4\dfrac{x+1}{x}=3\\ \dfrac{x+1}{x}=4^3\\ \dfrac{x+1}{x}=64\\ x+1=64x\\ 63x=1\\ x=\dfrac{1}{63}$