Question

How do you evaluate `log_9 729`?

Answer 1

When I write `log_(a)b=c`. I state explicitly that `a^c=b`

Explanation:

Typically we use sensible bases for logarithms such as `10` or `e`.

But with your problem we have `log_(9)729`.

And thus if `log_(9)729=x`, then `9^x=729=9^3`.

Clearly `log_(9)729` `=` `3`.

Answer 2

`3`

Explanation:

In log form, the question being asked is :

"To what power must 9 be raised to equal 729?"

It really is a huge advantage to know all the powers up to 1,000 by heart.

In this case you will recognize `729 = 9^3`

Therefore `log_9 729 = 3`

Without knowing this, you will have to do the whole log route...

`log_9 729 = x`
`9^x = 729`

#xlog9 = log729+

`x = (log729)/(log9) = 3`

It really is easier to just learn them!