Question

How do you solve `2log_5x=log_5 9`?

Answer 1

`x=3`

Explanation:

Remember the general relation:
`color(white)("XXX")log_b (a^c)=c * log_b (a)`

Therefore
`color(white)("XXX")2log_5(x)=log_5(x^2)=log_5(9)`

`color(white)("XXX")rarr x^2=9`

`color(white)("XXX")rarr x=3` (since the argument of `log` can not be negative)