Question

How do you solve `log_6(x+1)-log_6x=log_x 29`?

Answer 1

The proposed equation has not real solution.

Explanation:

`log_6(x+1)-log_6x=log_x 29` or
`log_e((x+1)/x)/log_e 6 = log_e 29/log_e x` or
`log_e x log_e((x+1)/x)= log_e6 log_e 29` or
`e^(log_e x log_e((x+1)/x)) = e^( log_e6 log_e 29)` or
`((x+1)/x)^(log_e x) = e^( log_e6 log_e 29)`

but

`((x+1)/x)^(log_e x)<((x+1)/x)^x < e`

and

`e^( log_e6 log_e 29)> e^6`

So, concluding, the proposed equation has not real solution.

Certainly, this is not a Precalculus question