Question

How to find the antilogarithm of In x = 2 to the nearest ten-thousandths

Answer 1

Given:

The expression is,

`\ln x=2`

But it is known that,

`\ln x=\log _ex`

Take antilog on both side,

`\begin{gathered} antilog_e\text{(log}_ex)=antilog_e(2) \\ \log ^(-1)_e(\log _ex)=e^2 \\ x=7.3890561 \\ \\ \end{gathered}`