Question 8: Huria is hiking up a mountain where the temperature \(T\) decreases by \(3\%\) (or a factor of \(0.97\)) for every \(100\) meters gained in altitude. The initital temperature \(T_i\) at sea level is \(20^o\ C\). Using the formula \(T=T_i \times 0.97^{\frac{h}{100}}\), calculate the temperature at an altitude \(h\) of \(500\) meters.
Solution:
It is given that
\[T=T_i \times 0.97^{\frac{h}{100}}\]
Put \(T_i=20\) and \(h=500\) in above equation, we have
\[
\begin{align*}
T&=20\times 0.97^{\frac{500}{100}}\\
T&=20\times 0.97^{5}
\end{align*}
\]
Taking logarithm on both sides, we have
\[
\begin{align*}
\log T&=\log\left( 2\times ( 0.97)^{5}\right)\\
&=\log 2+\log( 0.97)^{5}\\
&=\log 2+5\log 0.97\\
&=0.3010+5( -0.0132)\\
&=0.3010-0.066\\
&=0.235\\
\end{align*}
\]
Taking antilog on both sides, we have
\[
\begin{align*}
\text{anti-log}(\log x) &=\text{anti-log}( 0.235)\\
x&=1.72^o\\
\end{align*}
\]