# How do you evaluate log_9 (1/729)?

Nov 22, 2016

${\log}_{9} \left(\frac{1}{729}\right) = - 3$

#### Explanation:

In evaluating a log, read the expression as a question...

"To what power must 9 be raised to give $\frac{1}{729}$?"

[Nice to know $\rightarrow {9}^{3} = 729$]

${\log}_{9} \left(\frac{1}{729}\right)$

=${\log}_{9} \left(\frac{1}{9} ^ 3\right)$

=${\log}_{9} \left({9}^{-} 3\right)$

This actually answers the question, because there is an index of -3.

$\therefore {\log}_{9} \left(\frac{1}{729}\right) = - 3$

OR, using index form:

${\log}_{9} \left(\frac{1}{729}\right) = x \text{ " hArr" } {9}^{x} = \frac{1}{729}$

${9}^{x} = \frac{1}{9} ^ 3$

${9}^{x} = {9}^{-} 3$

$\therefore x = - 3$

Nov 22, 2016

Added note in support of EZ as pi

#### Explanation:

Using example:

$\textcolor{b r o w n}{\text{Important fact}}$

$\textcolor{b r o w n}{\text{Suppose we had "a^2" and wished to take logs of it. We would}}$ $\textcolor{b r o w n}{\text{have "log(a^2)". This has the same value as }}$
$\textcolor{b r o w n}{2 \times \log \left(a\right) = 2 \log \left(a\right)}$

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Consider log to base 10

Suppose we had log base 10 of 3 written as ${\log}_{10} \left(3\right)$. Normally written just as $\log \left(3\right)$

Set ${\log}_{10} \left(3\right) = x$
Another way of writing this is ${10}^{x} = 3$

Set ${\log}_{10} \left(10\right) = x$
Another way of writing this is ${10}^{x} = 10 \text{ " =>" } x = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider log to base e

Set $\to {\log}_{e} \left(3\right) = x$

This is a special log which is normally written as $\ln \left(3\right) = x$
Another way of writing this is ${e}^{x} = 3$

Set ${\log}_{e} \left(e\right) = x \to \ln \left(e\right) = x$
Another way of writing this is $\text{ "e^x=e" " =>" } x = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider log to base 9

Set ${\log}_{9} \left(\frac{1}{9} ^ 3\right) = x$

${\log}_{9} \left({9}^{- 3}\right) = x$

Write this as:

-3log_9(9)=x" "color(brown)( larr" From important note")

But ${\log}_{9} \left(9\right) = 1$ giving

$\left(- 3\right) \times 1 = x$

$x = - 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(blue)(ul(" So " log_b(b)=1)