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

08182014, 10:28 PM
Post: #12




RE: Math problem where graphing calculator may slow you down...
(08142014 05:12 PM)Thomas Klemm Wrote: similarly that \(e^{45} < 10^{20} < e^{225}\) Here's how you can show this with a 4banger. 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 

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