we need the same base in the logs
let log9 x = y
then 9^y = x
3^2y = x
and log3 x = 2y
so log9 x = 2log3 x
then
log(base 9)X + 3log(base 3)X = 14 becomes
2log3 + 3log3 - 14
5log3 = 14
log3 = 2.8
x = 3^2.8
= 21.674
how do i solve this logarithm
log(base 9)X + 3log(base 3)X = 14
1 answer