Question

How do you calculate `log_2 512`?

Answer 1

`log_2(512)= 9`

Explanation:

Notice that 512 is `2^9`.
`implies log_2(512) = log_2(2^9)`
By the Power Rule, we may bring the 9 to the front of the log.
`= 9log_2(2)`

The logarithm of a to the base a is always 1. So `log_2(2) = 1`
`=9`

Answer 2

the value of `log_(2)512=9`

Explanation:

we need to calculate `log_2(512)`
`512=2^9rArrlog_2(512)=log_2(2^9)`
`log_ab^n=nlog_ab` `rArrlog_(2)2^9=9log_(2)2`
since `log_(a)a=1rArrlog_(2)512=9`

Answer 3

`log_2 512 = 9" "` because `2^9=512`

Explanation:

Powers of numbers can be written in index form or log form.
They are interchangeable.

`5^3 = 125` is index form: It states that `5xx5xx5 = 125`

I think of log form as asking a question. In this case we could ask:

"Which power of `5` is equal to `125?`"
or
"How can I make `5` into `125` using an index?"

`log_5 125 =?`

We find that `log_5 125 = 3`

Similarly:
`log_3 81 =4" "` because `3^4 =81`
`log_7 343 = 3" "` because `7^3 =343`

In this case we have:

`log_2 512 = 9" "` because `2^9=512`

The powers of `2` are:

`1, 2,4,8,16,32,64,128,256,512,1024`

(From `2^0=1` up to `2^10 = 1024`)

There is a real advantage in learning all the powers up to `1000`, there are not that many and knowing them will make your work on logs and exponential equations SO much easier.