Question 205138
Please help me simplify the following:

{{{2LN(x-3)+LN(x+2)-6LN(x)}}}
<pre><font size = 4 color = "indigo"><b>
Use the principle {{{A*LN(B)=LN(B^A)}}} to rewrite
the first and last terms:

{{{LN(x-3)^2+LN(x+2)-LN(x^6)}}}

Use the principle: {{{LN(A)+LN(B)=LN(A*B)}}} to rewrite
the first two terms:

{{{LN((x-3)^2(x+2))-LN(x^6)}}}

Use the principle: {{{LN(A)-LN(B)=LN(A/B)}}} to rewrite
the expression:

{{{LN(((x-3)^2(x+2))/(x^6))}}}

=================================================

{{{drawing(200,50,-.01,3,-.2,.1, locate(0,0,
x(1-2x)^(-3/2)+(1-2x)^(-1/2))  )}}}

The smaller of those two exponents is {{{-3/2}}},
so we factor out {{{(1-2x)^(-3/2)}}}

{{{drawing(200,50,-.01,3,-.2,.1, locate(0,0,
(1-2x)^(-3/2)  ( x+(1-2x)^1 ))  )}}}

You may wonder where I got that {{{1}}} exponent.
When you factor one power of {{{(1-2x)}}} out of
another power of {{{(1-2x)}}}, you divide by 
subtracting exponents, and {{{(-1/2)-(-3/2)=-1/2+3/2=2/2=1}}}

So we can now erase that 1 exponent:

{{{drawing(200,50,-.01,3,-.2,.1, locate(0,0,
(1-2x)^(-3/2)  ( x+(1-2x) ))  )}}}

Removing the parentheses within parentheses:

{{{drawing(200,50,-.01,3,-.2,.1, locate(0,0,
(1-2x)^(-3/2)  ( x+1-2x ))  )}}}

{{{drawing(200,50,-.01,3,-.2,.1, locate(0,0,
(1-2x)^(-3/2)  ( 1-x) )  )}}}

Edwin</pre>