Question 1091577
If you were not given the answer to {{{log(1/2,16)}}}{{{"="}}}{{{"?"}}} , there is a way to figure out that the answer is {{{4}}} .
If you are given that statement {{{log(1/2,16)}}}{{{"="}}}{{{-4}}} , you can verify/prove that is true by showing that {{{(1/2)^(-4)}}}{{{"="}}}{{{16}}} .
 
FINDING THE ANSWER:
If you need to calculate logarithms with an cumbersome base {{{c}}} ,
you can convert them to a more helpful base {{{b}}} , by using the fact that
{{{log(c,x)}}}{{{"="}}}{{{log(b,x)/log(b,c)}}} .
That is a theorem that can be formally proven,
and as a formula is useful, and easy to remember.
Applied to this case, using {{{2}}} as your helpful base
{{{log("1 / 2",16)}}}{{{"="}}}{{{log(2,16)/log(2,"1 / 2")}}}{{{"="}}}{{{4/(-1)}}}{{{"="}}}{{{highlight(-4)}}} .
 
PROVING GIVEN STATEMENT:
Here is one way to prove that {{{(1/2)^(-4)}}}{{{"="}}}{{{16}}} :
The fraction {{{1/2}}} is {{{1}}} divided by {{{2}}} ,
and you know that with exponents,
{{{(a/b)^n=a^n/b^n}}} is true, as long as {{{b<>0}}} .
So, in this case,
{{{(1/2)^(-4)}}}{{{"="}}}{{{1^(-4)/2^(-4)}}}{{{"="}}}{{{1/2^(-4)}}} ,
and you know that by the definition of negative exponents
{{{2^(-4)}}}{{{"="}}}{{{1/2^4}}}{{{"="}}}{{{1/16}}} ,
so {{{(1/2)^(-4)}}}{{{"="}}}{{{1/"1 / 16"}}}{{{"="}}}{{{16}}} .
You know that 
{{{(1/2)^4=(1/2)*(1/2)*(1/2)*(1/2)=1/16}}} ,
and you should know that
{{{1/"1 / 16"=16}}}
 
If you can think of {{{1/2}}} only as a fraction,
you still would not that the way negative exponents are defined,
{{{(1/2)^(-4)}}}{{{"="}}}{{{1/"( 1 / 2 )"^4}}}