Consider the exponential equation $$1.1^x = 2$$ Let $N$ and $N+1$ be the two INTEGERS which bracket the solution (ie. $1.1^N<2$ and $1.1^{N+1}>2$). Then
$N$ =

Now consider the exponential equation $$1.1^y = 3$$ Let $N$ and $N+1$ be the two INTEGERS which bracket the solution (ie. $1.1^N<3$ and $1.1^{N+1}>3$). Then
$N$ =

Using the integers $N$ as approximations to the actual solutions $x$ and $y$ of the exponential equations above we find that the exponential equation $$2^z =3$$ has approximate solution $z\approx\frac{m}{n}$ where
$m$ =
and $n$ =
To see how good an approximation $\frac{m}{n}$ is to the actual solution we compute
$2^{\frac{m}{n}}$ =
and check how close it is to 3.

This problem illustrates John Napier's (1550-1617) approach to solving exponential equations and how he came to discover natural logarithms. Note that computing INTEGER powers of 1.1 can be done easily by hand, as multiplication by 1.1 requires only shifting and adding.

HINT FOR PARTS 3 AND 4: $(1.1^a)^b = 1.1^{ab}$