# How do you evaluate log_9 9?

Jun 20, 2018

$1$

#### Explanation:

$\text{using the "color(blue)"law of logarithms}$

•color(white)(x)log_b x=nhArrx=b^n

$\text{let } {\log}_{9} 9 = n$

$\text{then } 9 = {9}^{n}$

${9}^{1} = {9}^{n} \Rightarrow n = 1$

$\text{this is a standard result}$

•color(white)(x)log_b b=1

Jun 21, 2018

$1$

#### Explanation:

Given: ${\log}_{9} \setminus 9$.

Using the definition of logarithms, which states that:

if ${b}^{y} = x , \therefore {\log}_{b} \setminus x = y$

So, we get:

${\log}_{9} \setminus 9 = x$

${9}^{x} = 9$

${9}^{x} = {9}^{1}$

$\therefore x = 1$

Jun 21, 2018

${\log}_{9} 9 = 1$

#### Explanation:

We can use the logarithm rule

${\log}_{a} a = 1$

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

${\log}_{b} x = a$

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!