Question 1209756
Looks like AI can solve this type problem.  Give AI a few years and it'll be
able to solve any mathematics problem.  In the beginning, anybody could beat
AI in chess.  But the last time a human could beat AI in chess was in 2008.
It hasn't quite gotten there in mathematics. But give it time and it will!  

<pre>
[In the US, we use a DOT "." for a decimal point, and COMMAS "," to separate
digits in groups of three. It is just the opposite in your country. But nobody
separates digits in groups of three when calculating.]

Since the bulldozer drops in value by 20% each year means that each year, its
value is only 80% of what it was the year before.  So, we are talking about the
geometric sequence with first term a<sub>1</sub>=160,000, and common ratio
r=0.80, and its nth term is a<sub>n</sub> = v<sub>n</sub>. 

(a)
160,000, (0.80)(160000), (0.80)<sup>2</sup>(160000), (0.80)<sup>3</sup>(140000),...

{{{a[n]}}}{{{""=""}}}{{{a[1]r^(n-1)}}} 

{{{v[n]}}}{{{""=""}}}{{{160000(0.80)^(n-1)}}}

(b)
Since it will only take a few years, it's easier to do it this way
The 1st year the value is $160,000.
The 2nd year the value is $160,000(0.80) = $128,000.
The 3rd year the value is $128,000(0.08) = $102,400.
The 4th year the value is $102,000(0.80) = $81,920.
So the 4th year is the first year its value will be less than $100,000 all year.

But your teacher might expect you to use the formula. So you'd do it
this way to please your teacher:

{{{v[n]}}}{{{""=""}}}
{{{160000(0.80)^(n-1)}}}{{{""<""}}}{{{100000}}}
{{{160000*expr(0.80^n/0.80)}}}{{{""<""}}}{{{100000}}}
{{{200000*0.80^n}}}{{{""<""}}}{{{100000}}}
{{{0.80^n}}}{{{""<""}}}{{{0.5}}}
{{{ln(0.80^n)}}}{{{""<""}}}{{{ln(0.5)}}}
{{{n*ln(0.80)}}}{{{""<""}}}{{{ln(0.5)}}}
{{{n}}}{{{""<""}}}{{{ln(0.5)/ln(0.80)}}}
{{{n}}}{{{""<""}}}{{{ln(0.5)/ln(0.80)}}}
{{{n}}}{{{""<""}}}{{{3.10628372}}}

So the year after 3.10628372 years its value will be less than $100,000.
That means that the 4th year is the first year its value will be less 
than $100,000 all year.

Edwin</pre>