Question

How do you solve the equation `log_(9)81+log_9 1/9+log_9 3= log_9x`?

Answer 1

`x=243`

Explanation:

Since
`color(white)("XXX")log_b 1 = 0` for `AAb`
therefore
`color(white)("XXX")(log_9 1)/9 = 0`

and
`color(white)("XXX")log_9 81+(log_9 1)/9 +log_9 3`
is equivalent to
`color(white)("XXX")log_9 81 +log_9 3`
and by log addition:
`color(white)("XXX")= log_9 (81xx3)`

So we need to solve
`color(white)("XXX")log_9 243 = log_9 x`

but by observation this is simply
`color(white)("XXX")x=243`