Question 1178811
<pre>
I think you meant CUBE root, not SQUARE root, so use "root(3,5-x)" not sqrt.

{{{log((10*x^4*root(3,(5-x))/(3(x+6)^2)))}}}

That's the log of a quotient, so we use {{{log((a/b))=log((a))-log((b))}}}

{{{log((10x^4*root(3,5-x)))}}}{{{""-""}}}{{{log((3(x+6)^2))}}}

Those are both logs of products, so we use {{{log((abc))=log((a))+log((b))+log((c))}}} on the first log:

{{{log((10))+log((x^4))+log((root(3,(5-x))))}}}{{{""-""}}}{{{log((3(x+6)^2))}}}

Now we must be careful to put a big parentheses around the second log's
break-up since it's preceded by a minus:

{{{log((10))+log((x^4))+log((root(3,(5-x))))}}}{{{""-""}}}{{{(log((3))+log(((x+6)^2)))}}}

Now we can remove that big parentheses by changing the signs of the terms
inside the big parentheses:

{{{log((10))+log((x^4))+log((root(3,(5-x))))-log((3))-log(((x+6)^2))}}}

Now we write the cube root as the 1/3 power:

{{{matrix(2,1,"",log((10))+log((x^4))+log(((5-x)^(1/3)))-log((3))-log(((x+6)^2)))}}}

Now we use {{{log((a^b))=b*log((a))}}} and also log(10) = 1:

{{{1+4*log((x))+expr(1/3)*log((5-x))-log((3))-2*log((x+6))}}}

Edwin</pre>