How to solve for $k$ when the area about the $x$ axis and under the graph of the $f(x) = \frac1x$ from interval $x = [2, k]$ is equal to $\ln(4)$?

Question

What approach would be ideal in solving for a number $k$ when the area about the $x$ axis and under the graph of the function $f(x) = \frac1x$ from interval $x = [2, k]$ is equal to $\ln(4)$?

Answer 1

Notice

$$ \int\limits_{2}^k \frac{1}{x} dx = \ln 4 = 2 \ln 2$$

This is given. But

$$ \int\limits_{2}^k \frac{1}{x} dx = \ln k - \ln 2 $$

Hence,

$$ \ln k - \ln 2 = 2 \ln 2 \implies \ln k = 3 \ln 2 \implies \ln k = \ln 2^3 \implies k = 2^3 = 8 $$