I need help with Expanding logs

Question

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)

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

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

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

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

Please use * to denote multplication, or even better use MathJax to format your question. — , May 01 '14 at 00:07
@Trent I have formatted your question. Please double-check to ensure that I have correctly transcribed the expressions. — Neal, May 01 '14 at 00:11
The King will not complete all without seeing what has been attempted. — King Squirrel, May 01 '14 at 00:17
@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. — Patrick Da Silva, May 01 '14 at 00:34
The downvote was completely unjustified. I +1'ed to compensate... — Patrick Da Silva, May 01 '14 at 01:43

score 1 · Answer 1 · answered May 01 '14 at 00:30

1

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

answered May 01 '14 at 00:30

Ross Millikan

374,822

Thanks! So I see that the exponents apply first then the log rules. – Trent May 01 '14 at 00:31
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. – Ross Millikan May 01 '14 at 00:36
Thanks! How about the 3rd problem? The other two seem to be answered by your answer. – Trent May 01 '14 at 00:37
The cube root is just the $\frac 13$ power – Ross Millikan May 01 '14 at 00:38
Ah makes sense. Thanks! – Trent May 01 '14 at 00:39
@Trent consider mark the question as solved – ESCM Apr 27 '20 at 17:51

I need help with Expanding logs

1 Answers1