How do you evaluate `log_2 (1/2)`?

`log_color(red)(2)color(blue)( (1/2 ))`

Written in this form, it helps to read it as a question ...

"What power of `color(red)(2)` will give `color(blue)(1/2)?`

We should realise that `color(blue)(1/2)` is the reciprocal of `color(red)(2)`, and it can be written as `2^-1`

Log form and index form are inter-changeable.

`log_a b = c hArr a^c = b`

So, `log_color(red)(2)color(blue)( (1/2 )) = color(teal)(-1)" "hArr " " color(red)(2)^color(teal)(-1) = color(blue)(1/2)`

In the same way, can you answer the following instantly?

`log_3 27 ,color(white)(xx) log_5 625,color(white)(xx) log_7 49, color(white)(xx)log_2 32, color(white)(xx)log_8 2?`

In reverse order the answers are:

`1/3,color(white)(xxxx)5,color(white)(xxxx)2,color(white)(xxxx)4,color(white)(xxxx)3`

Let `log_2(1/2)=x`.......................(1)

Another way of writing this is:

`2^x=1/2`.......................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
AS an example compare to `5^3/5^4`

This is `5^(3-4) = 5^(-1) = 1/5`

So any value written as say `color(red)(z^(-1))` is the same as `1/z`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So by comparing to the example:

`2^x=1/2 -> x=-1`

`color(red)(2^(-1))=1/2` as in equation(2)

Thus equation(1) is

`log_2(1/2)=-1`