# If $(1.001)^{1259} = 3.52$ and $(1.001)^{2062} = 7.85$, then $(1.001)^{3321}= ?$

If $(1.001)^{1259} = 3.52$ and $(1.001)^{2062} = 7.85$, then $(1.001)^{3321}= ?$

what should be the approach in-order to get a solution without electronic aid?

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If you can't get electronic aid, then you can either use a mechanical computer or do the computation by hand. Another hint: $1259+2062=3321$ (done without electronic aid). And since in fact $1.001^{1259}\neq3.52$ (also checked by mere inspection), you can in actually derive any result you like (ex falso sequitur quodlibet: from a false hypothesis any conclusion can be drawn). – Marc van Leeuwen Feb 12 '12 at 13:51
@MarcvanLeeuwen: How did you arrive at $1.001^{1259}\neq3.52$ by inspection? $1.001^{1259}\approx3.51968$, so $3.52$ isn't far off at all. – Isaac Feb 12 '12 at 19:19
@Isaac: By the binomial formula, $1.001^{1259}$ has a digit $1$ at position $3777$ after the decimal point, while $3.52$ doesn't. So they differ. – Marc van Leeuwen Feb 12 '12 at 20:16
@MarcvanLeeuwen: Ahh, so the question would perhaps be better worded as "Since $(1.001)^{1259}\approx3.52$ and $(1.001)^{2062}\approx7.85$, $(1.001)^{3321}\approx ?$" – Isaac Feb 12 '12 at 20:19
@Isaac: Most certainly. – Marc van Leeuwen Feb 12 '12 at 20:20

$\begin{eqnarray} 1.001^{3321} &=& 1.001^{1259 + 2062} \\ &=& 1.001^{1259} \times 1.001^{2062}\end{eqnarray}$
$(1.001)^{3321}==(1.001)^{1259+2062}=>(1.001)^{1259}×(1.001)^{2062}$