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(log3x)2 + log3x2 + 1 = 0
On the second term use the rule
logban = n·logba
(log3x)2 + 2·log3x + 1 = 0
Let the letter w, or any letter other
than x, be such that
w = log3x
Then the above equation becomes
(w)2 + 2·w + 1 = 0
w2 + 2w + 1 = 0
Factoring
(w + 1)(w + 1) = 0
So the solutions w = -1 and w = -1 are
equal.
Now since w = log3x, then w = -1 becomes
log3x = -1
Now use the rule of logarithms that says:
logba = c can be rewritten as a = bc
to rewrite log3x = -1 as
x = 3-1
or x = 1/3
Edwin