Question

Log_6⁡(x+1)+log_6⁡x=1

Answer 1

This is pretty simple once you figure out the trick.

All you have to do is make the args common.

Meaning $log_{6}(x + 1) + log_{6}(x) = 1$ is nothing but $log_{6}(x(x + 1)) = 1$

Which is nothing but: $log_{6}(x^2 + x) = 1$.

Now using the scorpion tail method, this equation can be reframed to being: $6 = x^2 + x$. You bring 6 to the other side and it'll become: $x^2 + x - 6 = 0$

Factorise it and you get: $(x - 2) * (x + 3)$

Henceforth, the solution would be $2$ and $-3$. :D