Question
how do i solve this logarithm
log(base 9)X + 3log(base 3)X = 14
log(base 9)X + 3log(base 3)X = 14
Answers
Reiny
we need the same base in the logs
let log<sub>9</sub> x = y
then 9^y = x
3^2y = x
and log<sub>3</sub> x = 2y
so log<sub>9</sub> x = 2log<sub>3</sub> x
then
log(base 9)X + 3log(base 3)X = 14 becomes
2log<sub>3</sub> + 3log<sub>3</sub> - 14
5log<sub>3</sub> = 14
log<sub>3</sub> = 2.8
x = 3^2.8
= 21.674
let log<sub>9</sub> x = y
then 9^y = x
3^2y = x
and log<sub>3</sub> x = 2y
so log<sub>9</sub> x = 2log<sub>3</sub> x
then
log(base 9)X + 3log(base 3)X = 14 becomes
2log<sub>3</sub> + 3log<sub>3</sub> - 14
5log<sub>3</sub> = 14
log<sub>3</sub> = 2.8
x = 3^2.8
= 21.674