# Is there a way to answer this problem without using arithmetic at each step?

I was running through this mock high school math graduation test (try it out!) http://www.minnpost.com/data/2012/12/can-you-pass-mathematics-grad-test-high-school-students

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?

-
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

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. - 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$(0.75)^8=\frac{6561}{65536}$. - 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.
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 $(\frac{3}{4})^8\times 900$.