SOLUTION: how do you solve {{{16^(-2a-1)=1/2}}}

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Question 459009: how do you solve 16%5E%28-2a-1%29=1%2F2
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
16%5E%28-2a-1%29=1%2F2

Write 16 as 2⁴

%282%5E4%29%5E%28-2a-1%29=1%2F2

Multiply the exponents on the left to
remove the parentheses:

2%5E%28-8a-4%29=1%2F2

Multiply both sides by 2

2%2A2%5E%28-8a-4%29=2%2Aexpr%281%2F2%29

Add the understood 1 exponent of 2 on the left
to the other -8a-4 exponent of 2 on the left and 
get an exponent of -8a-4+1 or -8a-3

2%5E%28-8a-3%29=1

Replace the 1 on the right with 20
(That may seem strange but it causes both sides
to be powers of the same base 2. You are 
accustomed to replacing 20 by 1 but
not to replacing 1 by 20. So this
may be a first time with you).

2%5E%28-8a-3%29=2%5E0

Since the bases of exponents on each side of
the equation are the same positive number other
than 1, we may set the exponents equal:

So we set the exponents equal and drop the bases
of 2:

-8a-3=0

-8a=+3

a=-3%2F8

Edwin