Fully expand the logarithm ?

= Log[5](3x²/14) → you know that: Log(a/b) = Log(a) - Log(b)

= Log[5](3x²) - Log[5](14)

= Log[5](3 * x²) - Log[5](7 * 2) → you know that: Log(a * b) = Log(a) + Log(b)

= Log[5](3) + Log[5](x²) - { Log[5](7) + Log[5](2) }

= Log[5](3) + Log[5](x²) - Log[5](7) - Log[5](2) → you know that: Log[x^(a)] = a.Log(x)

= Log[5](3) + 2.Log[5](x) - Log[5](7) - Log[5](2)

log₅(3x² / 14)

First, let's get the prime factorization of the 14:

log₅[3x² / (7 * 2)]

Now, I'll turn the log of a product and quotient into the sum and differences of logs:

log₅(3) + log₅(x²) - log₅(7) - log₅(2)

Then the second term can have the exponent pulled out of the log:

log₅(3) + 2 log₅(x) - log₅(7) - log₅(2)

This is answer D.