Question 820859
<pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
log(x+3)=1-logx     |Note:  x > 0
log(x+3) + logx = 1
log x(x+3)= 1
   x(x+3) = 10^1
   x^2 + 3x - 10 = 0
  (x+5)(x-2) = 0  x = 2 (tossing out negative solution as x must be >0
CHECKING our answer***
 log 2*5 = 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]