How do you solve this log x base 3 plus log x3 base 9 equal 0?

log3(x) + log9(x^3)=0

for any t >0 logt(x) = ln(x)/ln(t)

so log9(x) = ln(x) / ln (9) = ln (x) / ( 2 * ln 3) = log3(x) /2 or log3(x) = 2 log9(x)

log9(x^n) = n * log9(x)

So log3(x) + log9(x^3) = log3(x) + 6 log3(x) = 7 log3(x)

7 log3(x) = 0 => log3(x) = 0 => x = 1