SOLUTION: Please find the unknown in the equation using steps log2 1056 = log2(2^a+5 plus 2^a) Please note that the 2 in the "log2 1056" is small: Please find the unknown in the equation

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Please find the unknown in the equation using steps log2 1056 = log2(2^a+5 plus 2^a) Please note that the 2 in the "log2 1056" is small: Please find the unknown in the equation      Log On


   



Question 133439: Please find the unknown in the equation using steps
log2 1056 = log2(2^a+5 plus 2^a)
Please note that the 2 in the "log2 1056" is small: Please find the unknown in the equation using steps
Thanks for answering the answer is correct:)
but how do I answer this quesiton with logs appearing in at least one of the steps?

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
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%282%5Ea%2B5+%2B+2%5Ea%29+
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%5E%28a%2B5%29+%2B+2%5Ea.
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%5E%28a%2B5%29+%2B+2%5Ea
1056+=+2%5Ea+%2A+2%5E5+%2B+2%5Ea
+1056+=+32+%2A+2%5Ea+%2B+2%5Ea+
+1056+=+33+%2A+2%5Ea
+32+=+2%5E5
a = 5