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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?

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  • $\begingroup$ You are correct. :-) $\endgroup$
    – rae306
    Aug 28, 2014 at 5:28
  • $\begingroup$ You are orrect and your steps are good :-) $\endgroup$ Aug 28, 2014 at 5:49

4 Answers 4

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

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

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Alternatively:

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

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Yes that is correct to my knowledge.

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  • $\begingroup$ 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. $\endgroup$ Aug 28, 2014 at 7:48
  • $\begingroup$ @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... $\endgroup$
    – user1729
    Aug 28, 2014 at 8:53
  • $\begingroup$ @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". $\endgroup$ Aug 28, 2014 at 9:01
  • $\begingroup$ @Jean-ClaudeArbaut Well, yes, but then the question would remain unanswered. $\endgroup$
    – user1729
    Aug 28, 2014 at 9:35
  • $\begingroup$ @user1729 Ok, I must admit, when I saw this answer in the review list, I didn't check the question :-) $\endgroup$ Aug 28, 2014 at 10:26

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