Math problem where graphing calculator may slow you down...

[Thomas Klemm]

(08-14-2014 05:12 PM)Thomas Klemm Wrote:  similarly that \(e^{45} < 10^{20} < e^{225}\)

Here's how you can show this with a 4-banger.
For this we want to calculate \(\log(e)\) where \(\log\) is the common logarithm (base 10) while \(\ln\) shall denote the natural logarithm.
Of course we know that \(\log(e)=\frac{\ln(e)}{\ln(10)}=\frac{1}{\ln(10)}\).
To calculate \(ln(10)\) we can use the same trick as before but turn it around and calculate the roots:

10
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
\(\sqrt{x}\)
− 1
× 1024 =

This gives us:
2.305174528

Now instead of \(45\times\log(e)\) we calculate \(\frac{45}{\ln(10)}\) and get:
19.52129844

Thus indeed \(e^{45}<10^{20}\) and analogous for \(10^{20}<e^{225}\).

Cheers
Thomas