<PRE><B>Question 1164820</B>
&lt;pre&gt;
ln{[(𝑥^15)(x-1)^1/2]/(3x-16)}=𝐴ln𝑥+𝐵ln(𝑥−1)+𝐶ln(3𝑥−16)

1)with the constant A=?
2)the constant B=?
3)and the constant C=?

{{{ln((x^15)matrix(2,1, &quot; &quot;, (x - 1)^(1/2))/(3x - 16)) = A*ln(x) + B*ln(x - 1) + C*ln(3x - 16)}}}

{{{ln((x^15)matrix(2,1, &quot; &quot;, (x - 1)^(1/2))/(3x - 16))}}} = &lt;font color = red&gt;&lt;font size = 4&gt;&lt;b&gt;Aln(x) + Bln(x - 1) + Cln(3x - 16)&lt;/font&gt;&lt;/font&gt;&lt;/b&gt;

{{{ln(x^15) + matrix(2,1, &quot; &quot;, ln(x - 1)^(1/2)) - ln(3x - 16)}}} = &lt;font color = red&gt;&lt;font size = 4&gt;&lt;b&gt;Aln(x) + Bln(x - 1) + Cln(3x - 16)&lt;/font&gt;&lt;/font&gt;&lt;/b&gt;

{{{15ln(x) + (1/2)ln(x - 1) - ln(3x - 16)}}} = &lt;font color = red&gt;&lt;font size = 4&gt;&lt;b&gt;Aln(x) + Bln(x - 1) + Cln(3x - 16)&lt;/font&gt;&lt;/font&gt;&lt;/b&gt;

&lt;font color = red&gt;&lt;font size = 4&gt;&lt;b&gt;&lt;font color = blue&gt;&lt;font size = 4&gt; 15&lt;/font&gt;&lt;/font&gt;ln(x) + {{{highlight_green(1/2)}}}ln(x - 1)&lt;font color = black&gt;&lt;font size = 4&gt; - 1&lt;/font&gt;&lt;/font&gt;ln(3x - 16)&lt;/font&gt;&lt;/font&gt;&lt;/b&gt;
 &lt;font color = red&gt;&lt;font size = 4&gt;&lt;b&gt;
 &lt;font color = blue&gt;&lt;font size = 4&gt; A&lt;/font&gt;&lt;/font&gt;ln(x)&lt;/font&gt;&lt;/font&gt; &lt;font color = green&gt;&lt;font size = 4&gt;+  B&lt;/font&gt;&lt;/font&gt;&lt;font color = red&gt;&lt;font size = 4&gt;ln(x - 1)&lt;/font&gt;&lt;/font&gt;&lt;font color = black&gt;&lt;font size = 4&gt; + C&lt;/font&gt;&lt;/font&gt;&lt;font color = red&gt;&lt;font size = 4&gt;ln(3x - 16)&lt;/font&gt;&lt;/font&gt;


Comparing coefficients above, we see that {{{highlight(matrix(3,3, A, &quot;=&quot;, 15, B, &quot;=&quot;, 1/2, C, &quot;=&quot;, - 1))}}}&lt;/pre&gt;</PRE>