Estimating the value of $\ln2$ using $e^3$ and $2^{10}$

Question

I found this question in an old MAT paper but I'm not getting very far.

You are given that $e^3$ is approximately $20$ and that $2^{10}$ is approximately $1000$. Using this information a student can obtain an approximate value for $\ln2$. which of the following is it?

a) $7/10$

b) $9/13$

c) $38/55$

d) $41/59$

Answer 1

$e^3 \approx 20 = (1000)^{\frac{1}{3}} \times 2 \approx 2^{\frac{10}{3}} \times 2 = 2^{\frac{13}{3}}$. So taking natural logarithms gives

$ 3= \frac{13 \ln 2 }{3}$ and solving gives $\ln 2 \approx \frac{9}{13}$.

[Just saw already answered in comments.]

Answer 2

$2^{10} \approx 1000 = 10^3$ taking the log of both sides $10\ln 2 \approx 3\ln 10.$

Also from the given information and taking the log of both sides $e^3 \approx 20\implies 3 \approx \ln 20 = \ln 2+\ln 10$

Substitute the value of $\ln 10$ from one equation into the other will let you solve for $\ln 2$

$\ln 10 \approx 3 - \ln 2\\ 10\ln 2 \approx 3(3-\ln 2)\\ 13\ln 2 \approx 9\\ \ln 2 \approx \frac {9}{13}$

Answer 3

We have the system

$$ \ln 2 + \ln 10 \approx 3\\ 10\ln 2-3\ln10 \approx 0 $$

solving for $\ln 2, \ln 10$ we have

$$ \ln2\approx \frac{9}{13}\\ \ln 10 \approx \frac{30}{13} $$