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

Aug 24, 2016

The proposed equation has not real solution.

#### Explanation:

${\log}_{6} \left(x + 1\right) - {\log}_{6} x = {\log}_{x} 29$ or
${\log}_{e} \frac{\frac{x + 1}{x}}{\log} _ e 6 = {\log}_{e} \frac{29}{\log} _ e x$ or
${\log}_{e} x {\log}_{e} \left(\frac{x + 1}{x}\right) = {\log}_{e} 6 {\log}_{e} 29$ or
${e}^{{\log}_{e} x {\log}_{e} \left(\frac{x + 1}{x}\right)} = {e}^{{\log}_{e} 6 {\log}_{e} 29}$ or
${\left(\frac{x + 1}{x}\right)}^{{\log}_{e} x} = {e}^{{\log}_{e} 6 {\log}_{e} 29}$

but

${\left(\frac{x + 1}{x}\right)}^{{\log}_{e} x} < {\left(\frac{x + 1}{x}\right)}^{x} < e$

and

${e}^{{\log}_{e} 6 {\log}_{e} 29} > {e}^{6}$

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

Certainly, this is not a Precalculus question