Question 850825
<pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
{{{log(6,x) + log(6,(x+9)) = 2}}}
{{{log(6, x(x+9)) = 2}}}
  6^2 = x(x+9)
  x^2 + 9x - 36 = 0
 (x+12)(x-3) = 0,  x = 3  
Note x = -12 is extraneous

*[tex \large\ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
*[tex \large\ \ log_bx + log_by = log_b(xy) ]