Question 1008878
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{{{log(3,(5z+5))}}} - {{{log(3,(z^2-1))}}} = {{{0}}}   -----> 

{{{log(3,(5z+5))}}} = {{{log(3,(z^2-1))}}}   ----->

{{{5z + 5}}} = {{{z^2-1}}}   ----->

{{{z^2 - 5z -6}}} = {{{0}}}   ----->   (factor)   ----->

(z-6)*(z+1) = 0   ----->

{{{z[1]}}} = 6, {{{z[2]}}} = -1.

The root {{{z[2]}}} doesn't suit the original equation (why ?).

The only solution of the original equation is z = 6.
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