



 
Sure Peter. As long as approximation is okay. Just graph it, ideally using log paper. The situation boils down to (0.9)^{x} = 0.1 Thus you can draw y = (0.9)^{x} on logarithmic graph paper and look for the x value where it crosses y = 0.1. So I just need one cycle semilog paper. I know I start at 0.9. The next two are easy: 0.81 and 0.729. I plot these as X marks on the graph, then draw a straight line carefully through them (an exponential on log paper is drawn as a straight line). I see this crosses the 0.1 on the y axis at x = 22, so that means 22 years is the answer to the question. No repetitive calculations (indeed no calculator), no logarithms (except what's built into the graph paper) and a pretty good approximation. In fact, using logs, I see that x = (log 0.1) / (log 0.9) = 21.85. Not bad. Hope this helps,
 


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