How do I use the properties of logarithms to simplify log4 (4x^2) for x > 0?

log₄(4x²)

The first thing to do is to change the log of a product to the sum of two logs:

log₄(4) + log₄(x²)

For the first term, you now have the same base of the log as you have inside of it, so that cancels out and it leaves the exponent (1):

1 + log₄(x²)

The second term can have the exponent pulled out of the log and multiplied by the log of only the base:

1 + 2 log₄(x)

I think this is the simplified form that the question is asking for as an answer.

= Log[4](4x²) → recall: Log[a](x) = Ln(x) / Ln(a) ← where a is the base

= Ln(4x²) / Ln(4) → you know that: Ln(ab) = Ln(a) + Ln(b)

= [Ln(4) + Ln(x²)] / Ln(4)

= [Ln(2²) + Ln(x²)] / Ln(2²) → you know that: Ln[x^(a)] = a.Ln(x)

= [2.Ln(2) + 2.Ln(x)] / 2.Ln(2)

= 2.[Ln(2) + Ln(x)] / 2.Ln(2)

= [Ln(2) + Ln(x)] / Ln(2) → to go further

= [Ln(2) / Ln(2)] + [Ln(x) / Ln(2)]

= 1 + [Ln(x) / Ln(2)] → to go further if you want it

= 1 + Log[4](x)

1. log(ab) = log(a) + log(b)

log4(4x^2) = log4(4) + log4(x^2)

2. log(a^b) = blog(a)

log4(4x^2) = log4(4) + 2log4(x)

3. log[base b])(b) = 1

log4(4x^2) = 1 + 2log4(x)

4. log[base b](x) = log[base c](x)/log[base c](b)

log4(4x^2) = 1 + 2log(x)/log(4)

from #2

log4(4x^2) = 1 + 2log(x)/log(2^2)

log4(4x^2) = 1 + log(x)/log(2)