# Question 52535

Dec 23, 2016

color(blue)(A~~$76126.2661" to 4 decimal places") $\textcolor{b l u e}{k \approx 0.03949 \text{ to 5 decimal places}}$#### Explanation: Given:$\text{ "$A(2.7)^(kt) = "land value}$

A(2.7)^(3k)=$112486.....................Equation(1) A(2.7)^(7k)=$131593....................Equation(2)
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$\textcolor{b l u e}{\text{Determine } k}$

$E q u a t i o n \left(2\right) \div E q u a t i o n \left(1\right)$

(cancel(Acolor(white)(.))(2.7)^(7k))/(cancel(Acolor(white)(.))(2.7)^(3k))=(cancel($)131593)/(cancel($)112486)

${\left(2.7\right)}^{4 k} = \frac{131593}{112486}$

$4 k \log \left(2.7\right) = \log \left(\frac{131593}{112486}\right)$

$k = \log \left(\frac{131593}{112486}\right) \div 4 \log \left(2.7\right)$

$\textcolor{b l u e}{k \approx 0.039488 \text{ to 6 decimal places}}$
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$\textcolor{b l u e}{\text{Determine A}}$

Substitute for $k$ in $E q u a t i o n \left(1\right)$

A=($112486)/(2.7)^(3k) color(blue)(A~~$76126.2661" to 4 decimal places")#