How do you evaluate  (e^0.24)^2?

Aug 20, 2016

Answer:

${\left({e}^{0.24}\right)}^{2} = 1.616$.

Explanation:

Let $x = {\left({e}^{0.24}\right)}^{2}$

$\therefore x = {e}^{\left(0.24\right) \left(2\right)} = {e}^{0.48}$

To find its numerical value, let us use Log-table (base 10).

$x = {e}^{0.48}$

$\Rightarrow \log x = \log {e}^{0.48} = 0.48 \log e$

$\Rightarrow \log x = \left(0.48\right) \left(0.4343\right) = 0.208464 \cong 0.2085$

$\Rightarrow x = A n t i \log 0.2085 = 1.616$

Hence, ${\left({e}^{0.24}\right)}^{2} = 1.616$.