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I was running through this mock high school math graduation test (try it out!)

Question 6 gave me pause:

Dan bought a new computer for $900. Each year, the value of the computer decreased by 25% of the previous year's value. At this rate, what can Dan expect the approximate value of the computer to be after 8 years?

I can solve this easily enough by looping through the calculation for each year's deprecation. Programming has made me lazy, I let loops handle the arithmetic for me. But when solving by hand it was tedious. There must be a better way! Here's what I know: there is a rate of change which is constant at 25% / year, and we know the initial value at $900, and how long the change will be applied, 8 years, but I'm fuzzy on how to put it all together to get a formula.

My college calculus knowledge has imploded on itself, leaving little trace. The Wikipedia page on derivatives is causing my vision to blur. Where do I start to reason my way towards an intelligently-calculated answer?

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This is a little off-topic, but I began reading some of the questions in that Minn. GRAD test: I weep for the mathematical level of those high school kids. Hopefully that was an exam for the lowest level and there are higher ones... – DonAntonio Jan 31 '13 at 2:56
up vote 3 down vote accepted

Each year the value of the computer is $\frac34$ of its value the previous year, so after $8$ years the value is $$900\cdot\left(\frac34\right)^8$$ dollars, or $\$90.10$ to the nearest cent.

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I'm almost certain it is not 60.07 – Calvin Lin Jan 30 '13 at 23:51
@Calvin: True. I wonder what on earth I punched into the calculator. – Brian M. Scott Jan 30 '13 at 23:52

If $a$ is the value this year, then the value next year is $(0.75)a$. So the value after $2$ years is $(0.75)(0.75)a$. After $3$ years it is $(0.75)(0.75)(0.75)a$, and so on.

So the value after $8$ years is $(0.75)^8 a$.

To evaluate $(0.75)^8$ on a simple calculator, square $0.75$, square the result, then square again, and it's done. Or else one can use the $y^x$ key. It's even practical to do it by hand. For $0.75=\frac{3}{4}$. Note that $3^8=(3^4)^2=81^2=6561$. Also, $4^8=2^{16}=65536$. So $(0.75)^8=\frac{6561}{65536}$.

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This is almost magic (i.e. unique to the values).

We're looking for $0.75^8$. The trick here is to know that $\log 0.75 \approx -0.125$. As such, $\log (0.75)^8 = 8 \times \log 0.75 \approx -1$. Thus, it will be worth about 1/10 of it's original price, i.e. $90.

So, to obtain $\log 0.75$, you can use $\log 0.75 = \log \frac {3}{2^2} \approx 0.477 - 2 \times 0.301 $. It is useful to know what $\log 2, \log 3$ is approximately equal to.

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After the first year, the value is $\frac{3}{4}\times 900$. After another year, it is $\frac{3}{4}\times\frac{3}{4}\times 900$. Continuing this way, one sees that after eight years, the value is $(\frac{3}{4})^8\times 900$.

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Yay, + 15K!!!!! – amWhy Jan 31 '13 at 0:03

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