# I need help with Expanding logs

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. – user61527 May 1 '14 at 0:07
• @Trent I have formatted your question. Please double-check to ensure that I have correctly transcribed the expressions. – Neal May 1 '14 at 0:11
• The King will not complete all without seeing what has been attempted. – King Squirrel May 1 '14 at 0:17
• @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 1 '14 at 0:34
• The downvote was completely unjustified. I +1'ed to compensate... – Patrick Da Silva May 1 '14 at 1:43

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
• 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 1 '14 at 0:36
• The cube root is just the $\frac 13$ power – Ross Millikan May 1 '14 at 0:38