Question 936216
log (k+2) + log (k-1) = 1   Note: 10^1 = 10
(k+2)(k-1) = 10
k^2 + k - 12 = 0
{k+4)(k-3)= 0
k = -4, 3
.......
*[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]***