How do you evaluate #log_9 9#?

3 Answers
Jun 20, 2018

#1#

Explanation:

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)log_b x=nhArrx=b^n#

#"let "log_9 9=n#

#"then "9=9^n#

#9^1=9^nrArrn=1#

#"this is a standard result"#

#•color(white)(x)log_b b=1#

Jun 21, 2018

#1#

Explanation:

Given: #log_9\9#.

Using the definition of logarithms, which states that:

if #b^y=x,:.log_b\x=y#

So, we get:

#log_9\9=x#

#9^x=9#

#9^x=9^1#

#:.x=1#

Jun 21, 2018

#log_9 9=1#

Explanation:

We can use the logarithm rule

#log_aa=1#

Since the base is the same as the thing we're taking the logarithm of, this evaluates to #1#. Let's make sure this makes sense.

In general, we know if we have

#log_bx=a#

Then this can be rewritten as

#b^a=x#

In our scenario, our #b=9#, and so does our #x#. Plugging in, we get

#9^a=9#

We can rewrite this as

#9^a=9^1#

Since the bases are equivalent, so are the exponents. Thus,

#a=1#

This confirms for us that

#log_9 9=1#

Hope this helps!