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

1 Answer
Cesareo R.
Aug 24, 2016

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

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