Given that 32√2 = 2^a. Find a?

Note: 32 = 2^5, and √2 = 2^(1/2).

Hence: 32√2 = 2^5 * 2^(1/2) = 2^(5 + 1/2) = 2^(11/2).

If: 2^(11/2) = 2^a ---> a = 11/2.


Answer: 11/2.

The answer is as follows:

2^(5.5) = 2^a
a = 5.5

32√2 = 2^a
32√2 = 2^11/2
a = 11/2

log [ 32 √2 ] = a log 2

log [ 32 √2 ] / log 2 = a

32√2 = 2^a
2⁵√2 = 2^a

You don't really need logs to solve this, just a little logic. Since the right side is a power of 2, we can write the left side to be a power of 2 as well. Let's put everything under the square root:

√(2⁵ * 2⁵ * 2) = 2^a
√(2¹¹) = 2^a

2^(11/2) = 2^a

Now that we have two values that are equal, and both bases are the same, the exponents must be the same:

a = 11/2

a = log₂(32√2) = log₂32 + log₂√2 = 5 + 1/2 = 5½

32 = 2*2*2*2*2 = 2^5
sqrt(2) = 2^0.5
Hence 32*sqrt(2) = 2^5.5
Hence a = 5.5 <<<

32√2 = 2^a
2⁵ 2^½ = 2^a
2^(5+½) = 2^a
2^(11/2) = 2^a
a = 11/2

Take base 2 logarithms

log_2 (32 √2) = a
log_2(32) + log_2(√2) = a .... log of product property
5 + 1/2 = a

So, a is 5.5 or 11/2. I like 11/2 because it looks exact, while 5.5 looks "good to one decimal place".

PS: Yes, you do need to know that 2^5 = 32, and that √x = x^(1/2) to take those logarithms exactly.