Due to ever-changing technology, a new XYZ Smartphone decreases in value 20% each year. 1. How much will this $1000 phone be worth in 2 years? 2. How long until it is worth less than 10% of it's original price?

Answer:1.  $6402. About 10.3 years laterStep-by-step explanation:This is a compound decay problem. The formula is[tex]F=P(1-r)^t[/tex]WhereF is the future amountP is the initial amountr is the rate of decrease (in decimal), andt is the time in yearsQuestion 1:We want to find F after 2 years of a phone initially costing 1000. So,P = 1000r = 20% or 0.2t = 2plugging into the formula, we solve for F:[tex]F=P(1-r)^t\\F=1000(1-0.2)^2\\F=1000(0.8)^2\\F=640[/tex]The phone is worth $640 after 2 yearsQuestion 2:We want to find when will the phone be worth 10% of original. 10% of 1000 is 0.1 * 1000 = 100So, we want to figure this out for future value of 100, so F = 100We know, P = 1000 r = 0.2 and t is unknown.Let's plug in and solve for t (we need to use logarithms):[tex]F=P(1-r)^t\\100=1000(1-0.2)^t\\100=1000(0.8)^t\\\frac{100}{1000}=0.8^t\\0.1=0.8^t\\ln(0.1)=ln(0.8^t)\\ln(0.1)=t*ln(0.8)\\t=\frac{ln(0.1)}{ln(0.8)}\\t=10.32[/tex]So, after 10.32 years, the phone would be worth less than 10% of original value.