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

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?

• You are correct. :-) Aug 28, 2014 at 5:28
• You are orrect and your steps are good :-) Aug 28, 2014 at 5:49

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 $2n \log_2 4 = 2n \cdot 2 = 4n$.

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

Alternatively:

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

Yes that is correct to my knowledge.

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. Aug 28, 2014 at 7:48
• @Jean-ClaudeArbaut Well, it does answer the question. The question is "Is this correct or am I completely wrong?". So "yes" is a valid answer, and is what the OP is after. Could this question be improved? Of course! But it does answer the question... Aug 28, 2014 at 8:53
• @user1729 It's not an answer, in the sense that it should be a comment. You don't need a full answer to say "yes". Aug 28, 2014 at 9:01
• @Jean-ClaudeArbaut Well, yes, but then the question would remain unanswered. Aug 28, 2014 at 9:35
• @user1729 Ok, I must admit, when I saw this answer in the review list, I didn't check the question :-) Aug 28, 2014 at 10:26