You buy a new computer for $2,100. the computer decreases by 50% annually.when will the computer have a value less than $100?

To determine when the computer will have a value of less than $100, we apply the concept of exponential decay based on the percentage decrease. In this case, the computer decreases by 50% annually. This means that every year, the value of the computer will be half of what it was the previous year.

Let's represent the initial value of the computer, $2,100, with V0. Let V represent the value of the computer after n years.

The formula for the value of the computer after n years is:

V = V0 * (0.5)^n

Here, 0.5 represents the 50% reduction each year (because 50% of the original amount would be remaining, so we multiply by 0.5, which is the same as dividing by 2).

To find out when the computer will have a value less than $100, we set V to be less than $100 and solve for n:

$100 > $2,100 * (0.5)^n

To solve for n, we can take the natural logarithm (or logarithm base 10, or in fact any logarithm as long as we are consistent on both sides) on both sides of the inequality:

ln($100) > ln($2,100 * (0.5)^n)

The logarithm of a product is the sum of the logarithms, so we can rewrite the right side of the inequality:

ln($100) > ln($2,100) + ln((0.5)^n)

Since ln((0.5)^n) is n * ln(0.5), the equation becomes:

ln($100) > ln($2,100) + n * ln(0.5)

Now we isolate n:

n > [ln($100) - ln($2,100)] / ln(0.5)

Plug in the values and use a calculator to find n:

n > [ln(100) - ln(2100)] / ln(0.5)

n > [-4.60517019 - 7.649753813] / -0.693147181

n > 3.044583624 / 0.693147181

n > 4.394

Since n represents full years, we can only pass full year increments, so rounding up to the next full year (you can't have a fraction of a year when counting elapsed time in full years), we get:

n > 5

So, the computer will have a value of less than $100 in more than 4 full years, which means it will be in the 5th year.