If it is known that log_2a+log_2b≥6, then the least value that can be taken on by a+b is:

1969 AMC 12 Problem 25

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25.

If it is known that

\log_2a+\log_2b\ge6,

then the least value that can be taken on by

a+b

is:

2\sqrt6

6

8\sqrt2

16

none of these

Combine the logarithms to bound

ab

Apply

a+b\ge2\sqrt{ab}

The logarithm condition gives

\log_2(ab)\ge6,

ab\ge64.

By AM-GM,

a+b\ge2\sqrt{ab}\ge2\sqrt{64}=16.

Equality is attained at

a=b=8.

Therefore, the correct answer is D.