How do you evaluate #log_9 729#?

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Answer 1 · 2016-08-07T13:35:09

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

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 · 2016-08-07T18:42:48

#3#

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!