Question 1103999
2. The first three terms of a geometric sequence 
are ln x^16, ln x^8, ln x^4, for x>0. 
a. Find the common ratio. 
<pre><font size=4><b>
a<sub>1</sub> = ln(x<sup>16</sup>) = 16&#8729;ln(x) 
a<sub>2</sub> = ln(x<sup>8</sup>) = 8&#8729;ln(x)
a<sub>3</sub> = ln(x<sup>4</sup>) = 4&#8729;ln(x)

r = a<sub>2</sub>/a<sub>1</sub> = 8&#8729;ln(x)/16&#8729;ln(x) = 1/2
</pre></font>
b.
<pre><font size=4><b>
{{{sum((2^(5-k)),k=1,infinity)*ln(x)}}}{{{""=""}}}{{{64}}}

First we find:

{{{sum((2^(5-k)),k=1,infinity)}}}{{{""=""}}}{{{a[1]/(1-r^"")}}} 

a<sub>1</sub> = 2<sup>5-1</sup> = 2<sup>4</sup> = 16
a<sub>2</sub> = 2<sup>5-2</sup> = 2<sup>3</sup> = 8

r = a<sub>2</sub>/a<sub>1</sub> = 8/16 = 1/2

{{{sum((2^(5-k)),k=1,infinity)}}}{{{""=""}}}{{{a[1]/(1-r^"")}}}{{{""=""}}}{{{16^""/(1-1/2)}}}{{{""=""}}}{{{16^""/(1/2)}}}{{{""=""}}}{{{16(2/1)}}}{{{""=""}}}{{{16(2)}}}{{{""=""}}}{{{32}}}

Then 

{{{sum((2^(5-k)),k=1,infinity)*ln(x)}}}{{{""=""}}}{{{64}}}

becomes

{{{(32)*ln(x)}}}{{{""=""}}}{{{64}}}

{{{ln(x)}}}{{{""=""}}}{{{64/32}}}

{{{ln(x)}}}{{{""=""}}}{{{2}}}

raise e to the powers of both sides

{{{e^ln(x)}}}{{{""=""}}}{{{e^2}}}

{{{x}}}{{{""=""}}}{{{e^2}}}

Edwin</pre></font></b>