How do you calculate #log_2 512#?

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#

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#

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.