Question 817168
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
log(3)k + log(3)k-2 = 1          |  *[tex \large\ \ log_bx + log_by = log_b(xy) ]
{{{log(3,(k(k-2))) = 1}}}       |*[tex \large\ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
{{{k(k-2) = 3^1}}}
  k(k-2) = 3
k^2 - 2k - 3 = 0
(k-3)(k+1) + 0
    k = 3    |Tossing out negative solution  k = -1
 CHECKING our answer***
{{{log(3,3) + log(3,1) =  1 + 0 =  1 }}} 
  
*[tex \large\ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
*[tex \large\ \ nlog_bx = log_b(x^n) ]
*[tex \large\ \ log_bx + log_by = log_b(xy) ]
*[tex \large\ \ log_bx - log_by = log_b(x/y) ]
*[tex \large\ \ log_b1 = 0]
*[tex \large\ \ log_bb = 1]