Question

How do you evaluate `log_9 9`?

Answer 1

`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`

Answer 2

`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`

Answer 3

`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!