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

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

"To what power must 9 be raised to give `1/729`?"

[Nice to know `rarr 9^3= 729`]

`log_9 (1/729)`

=`log_9 (1/9^3)`

=`log_9 (9^-3)`

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

`:. log_9 (1/729) = -3`

OR, using index form:

`log_9 (1/729) = x " " hArr" " 9^x = 1/729`

`9^x = 1/9^3`

`9^x = 9^-3`

`:. x = -3`

Using example:

`color(brown)("Important fact")`

`color(brown)("Suppose we had "a^2" and wished to take logs of it. We would")` `color(brown)("have "log(a^2)". This has the same value as " )`
`color(brown)(2xxlog(a) = 2log(a))`

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Consider log to base 10

Suppose we had log base 10 of 3 written as `log_10(3)`. Normally written just as `log(3)`

Set `log_10(3)=x`
Another way of writing this is `10^x=3`

Set `log_10(10)=x`
Another way of writing this is `10^x=10" " =>" " x=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider log to base e

Set `-> log_e(3)=x`

This is a special log which is normally written as `ln(3)=x`
Another way of writing this is `e^x=3`

Set `log_e(e)=x -> ln(e)=x`
Another way of writing this is `" "e^x=e" " =>" " x=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider log to base 9

Set `log_9(1/9^3) = x`

`log_9(9^(-3))=x`

Write this as:

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

But `log_9(9)=1` giving

`(-3)xx1=x`

`x=-3`
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`color(blue)(ul(" So " log_b(b)=1)`