Question 859225
<pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
1/3 log(3X+1)=2
*[tex \large\ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
*[tex \large\ \ nlog_bx = log_b(x^n) ]

{{{root(3,(3x+1)) = 100}}}
    3x+ 1 = 1000000
    3x = 999999
     x = 333333