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#