After simplification, the expression log2 3 โˆ’ 3 log2 ๐‘ฅ + 1/2 log2 (๐‘ฅ + 1) can be reduced to?

Using logarithm properties, we can simplify the given expression as follows:

log2 3 โˆ’ 3 log2 ๐‘ฅ + 1/2 log2 (๐‘ฅ + 1)

= log2 3 โˆ’ log2 ๐‘ฅ^3 + log2 (๐‘ฅ + 1)^(1/2)

= log2 (3/(๐‘ฅ^3)) + log2 (โˆš(๐‘ฅ + 1))

= log2 [(3/(๐‘ฅ^3)) * โˆš(๐‘ฅ + 1)]

Therefore, after simplification, the expression log2 3 โˆ’ 3 log2 ๐‘ฅ + 1/2 log2 (๐‘ฅ + 1) can be reduced to log2 [(3/(๐‘ฅ^3)) * โˆš(๐‘ฅ + 1)].

I thought it was simpler at the beginning :)

Apologies for any confusion. Let's try to simplify the expression further.

Starting with the expression log2 3 โˆ’ 3 log2 ๐‘ฅ + 1/2 log2 (๐‘ฅ + 1), we can apply logarithm properties to simplify it.

First, we can use the power rule of logarithms which states that log(base a) b^c = c * log(base a) b.

Therefore, we can rewrite the expression as:

log2 3 โˆ’ log2 (๐‘ฅ^3) + (1/2) * log2 (๐‘ฅ + 1)

Now, we can apply another logarithm property which states that log(base a) b - log(base a) c = log(base a) (b/c).

Using this property, we can combine the first two terms as:

log2 (3/(๐‘ฅ^3)) + (1/2) * log2 (๐‘ฅ + 1)

Finally, there isn't any further simplification possible without additional information or algebraic manipulation. Therefore, after simplification, the expression remains as log2 (3/(๐‘ฅ^3)) + (1/2) * log2 (๐‘ฅ + 1).