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I know the log rules for expanding but I am not sure how to expand these difficult ones: (also the x's here depending upon the context are multiplication)

$$\log_4(x^4yz)^2$$

$$\log_3 (((6\times 5)^2)/11)^2$$

$$\log_2(d \sqrt[3]{abc})$$

$$\log ((3\times 5)/8^3)^2$$

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  • $\begingroup$ Please use * to denote multplication, or even better use MathJax to format your question. $\endgroup$ – user61527 May 1 '14 at 0:07
  • $\begingroup$ @Trent I have formatted your question. Please double-check to ensure that I have correctly transcribed the expressions. $\endgroup$ – Neal May 1 '14 at 0:11
  • $\begingroup$ The King will not complete all without seeing what has been attempted. $\endgroup$ – King Squirrel May 1 '14 at 0:17
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    $\begingroup$ @Trent : Note that your notation is a bit ambiguous ; you seem to write $\log_4(x^4yz)^2$ to mean $\log_4( (x^4yz)^2 )$, which is really not the same as $(\log_4(x^4yz))^2$. Without a context, I would've read you wrong. $\endgroup$ – Patrick Da Silva May 1 '14 at 0:34
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    $\begingroup$ The downvote was completely unjustified. I +1'ed to compensate... $\endgroup$ – Patrick Da Silva May 1 '14 at 1:43
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If you know the log rules, you also need the rules for exponents, then just apply them. For example $$\log_4(x^4yz)^2=\log_4(x^8y^2z^2)=\log_4(x^8)+\dots=8\log_4x+\dots$$ The same approach works on all the rest

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  • $\begingroup$ Thanks! So I see that the exponents apply first then the log rules. $\endgroup$ – Trent May 1 '14 at 0:31
  • $\begingroup$ You could do it the other way, as well $$\log_4(x^4yz)^2=2\log_4(x^4yz)=$$ Both are legal, sometimes one is more helpful, sometimes the other. $\endgroup$ – Ross Millikan May 1 '14 at 0:36
  • $\begingroup$ Thanks! How about the 3rd problem? The other two seem to be answered by your answer. $\endgroup$ – Trent May 1 '14 at 0:37
  • $\begingroup$ The cube root is just the $\frac 13$ power $\endgroup$ – Ross Millikan May 1 '14 at 0:38
  • $\begingroup$ Ah makes sense. Thanks! $\endgroup$ – Trent May 1 '14 at 0:39

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