Question 41178
{{{log( 8, (n-3)) +  log( 8, (n + 4)) = 1}}}
{{{log( 8, (n-3)(n+4)) = 1}}}
{{{(n-3)(n+4) = 8^1}}}
{{{n^2 + n - 12 = 8}}}
{{{n^2 + n - 20 = 0}}}
{{{(n + 5)(n - 4) = 0}}}
{{{n = -5}}} or {{{n = 4}}}
You can not take a logarithm of a negative value, so negative five is an extraneous solution.
Answer: {{{n = 4}}}