Hi,
Are there alternative ways of solving this equation? Could logarithms be used?
$\displaystyle a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
$\displaystyle a^{60} = k$ , where $\displaystyle k = 2.044$
$\displaystyle 60\log{a} = \log{k}
$
$\displaystyle \displaystyle \log{a} = \frac{\log{k}}{60}$
$\displaystyle \displaystyle a = e^{\frac{\log{k}}{60}}$
$\displaystyle a \approx 1.012$
note that $\displaystyle \displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}$