Simplifying/solving a logarithm $\log_24^{2n}$

Question

Need help with simplifying this logarithm.

$$\log_24^{2n}$$

Would I just pull the 2n to the front:

$$2n*\log_24$$

So it would simplify to $$4n$$

Is this correct or am I completely wrong?

Answer 1

Remember that $\log a^b = b \log a$. So $\log_2 4^{2n} = 2n\log_2 4$.

And you know that $\log_2 4 = 2$. If you ever get confused, just remember that $\log_{10} 1000 = 3$ because $10^3 = 1000$. So in this case, $\log_2 4 = 2$ because $2^2 = 4$.

So you have $2n \log_2 4 = 2n \cdot 2 = 4n$.

Answer 2

$\log_2 4^{2n}=\log_22^{2\cdot2n}=\log_22^{4n}$ (since $4=2^2)$.

Now ask yourself the following question: to what power do I have to raise $2$ to get $2^{4n}$? That's $4n$, so $\log_2 4^{2n}=4n$.

You are correct. :)

Answer 3

Alternatively:

$\log(4^2n) / \log(2) = 2n* [\log(2) + \log(2)] / \log(2) = 2n*(1+1) = 4* n$

Answer 4

Yes that is correct to my knowledge.