Question

How do you solve `log_6 u–log_6 9=2`?

Answer 1

The result should be `u=324`

You start using a property of logarithms where you have:
`log_a(x)-log_a(y)=log_a(x/y)`

and the basic definition of logarithm: `log_a(x)=b` `->` `x=a^b`

And using these in your example:

`log_6(u/9)=2` and so:
`u/9=6^2`
`u=9*6^2`
`u=324`