Question 205650
Assume x>0 and 3x-1>0. Rewrite the following expression as a single logarithm. 
<pre><font size = 4 color = "indigo"><b>
{{{2ln(x)-(1/2)ln(9x^2-1)+ (1/2)ln (3x+1) }}}

Use the fact that {{{A*ln(B) = ln(A^B)}}} on each of the
three terms:

{{{ ln(x^2)-ln(9x^2-1)^(1/2)+ ln (3x+1)^(1/2)}}}

Use the fact that {{{ln(A)-ln(B)=ln(A/B)}}} on the first
two terms:

{{{ ln((x^2)/(9x^2-1)^(1/2))+ ln (3x+1)^(1/2)}}}

Use the fact that {{{ln(A)+ln(B)=ln(A*B)}}} 

{{{ ln((x^2/(9x^2-1)^(1/2))(3x+1)^(1/2))}}}

You actually have it as a single logarithm at this step.
However, your teacher probably wants you to simplify it.  

Put the second factor over 1 so you will have two
fractions to multiply:

{{{ ln((x^2/(9x^2-1)^(1/2))(((3x+1)^(1/2))/1))}}}

Indicate the multiplication of tops and bottoms of 
those fractions:

{{{ ln(  (x^2(3x+1)^(1/2))  /((9x^2-1)^(1/2)))}}}

Factor {{{9x^2-1}}} as {{{(3x-1)(3x+1)}}}

{{{ ln(  (x^2(3x+1)^(1/2))  /(((3x-1)(3x+1))^(1/2)))}}}

Use the fact that {{{(A*B)^N=A^N*B^N}}} to rewrite the
denominator:

{{{ ln(  (x^2(3x+1)^(1/2))  /((3x-1)^(1/2)(3x+1)^(1/2)))}}}

Cancel:

{{{ ln(  (x^2*cross((3x+1)^(1/2)))  /((3x-1)^(1/2)cross((3x+1)^(1/2))))}}} 

{{{ ln(  x^2  /((3x-1)^(1/2)))}}}

Change the {{{1/2}}} power to a square root:

{{{ ln(  x^2  /sqrt(3x-1))}}}

Edwin</pre>