Question 133439
There have been a bunch of problems like this one recently. You might want to check the 'recently solved' before posting a problem that is already solved several times.

{{{log2 1056 = log2(2^a+5 + 2^a) }}}
In tis problem, you can elininate the Log2 from both sides. You can just 'know that since a Log2 on the left is equal to the Log2 on the right, then the values are the same. Thus {{{1056 = 2^(a+5) + 2^a}}}. 

If you want to go a different way, take both sides of the equation and use the values there as exponents on the power of 2. So 2^(log2 1056) = 2^(2^a+5 + 2^a).
That becomes 1056 Log2 2 = (2^(a+5) + 2^a) Log2 2. Since Log2 2 = 1, you get to the same equation as above.

{{{1056 = 2^(a+5) + 2^a}}}
{{{1056 = 2^a * 2^5 + 2^a}}}
{{{ 1056 = 32 * 2^a + 2^a }}}
{{{ 1056 = 33 * 2^a}}}
{{{ 32 = 2^5}}}
a = 5