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$\log_{10}x \approx x^{ \frac{1}{2^{19}}}227695 - 227695$

I plotted the graphs of both $\log_{10}x$ and $x^{ \frac{1}{2^{19}}} 227695 - 227695$

And found that they almost overlapped.

Can you give a explanation of this or can you find how this approximation is approached.

kurama
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    domain? It is certainly nowhere close if $x\approx \underbrace{10^{10^{10^{10^{10^{\dots}}}}}}_{10^{10}}$, for example. – user10354138 Jun 13 '19 at 14:38
  • I am talking about the value of $log_{10} x$ – kurama Jun 13 '19 at 14:44
  • but they are nowhere close. $\log_{10}x$ is the same form with $10^{10}-1$ underbrace, which is way smaller than $10^{10^{\dots}/2^{19}}$. – user10354138 Jun 13 '19 at 14:47
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    Try plugging in $x = 2^{19}$. You get $\log_{10} x \approx (2^{19})^\frac{1}{19} \cdot 227695 - 227695 = 2 \cdot 227695 - 227695 = 227695$. This is a far cry from the correct value, which is approximately $6$. This example shows that the approximation is not really a very good one. – John Hughes Jun 13 '19 at 14:47
  • @JohnHughes its $x^{ \frac{1}{2^{19}} }$ not $x^{\frac{1}{19}}$ – Rkb Jun 13 '19 at 14:53
  • @JohnHughes and who upvoted the comment -_- – Rkb Jun 13 '19 at 14:55
  • OK. Try plugging in $x = 2^{2^{19}}$. You get $\log_{10} x \approx (2^{2^{19}})^\frac{1}{2^{19}} \cdot 227695 - 227695 = 227695.$ The correct answer is $2^{19} \log_{10} 2 \approx 157826$. Still pretty darned wrong. And (no surprise), the error grows as $x$ grows, because...well, because $\log$ isn't a polynomial. – John Hughes Jun 13 '19 at 14:58
  • When you write "I plotted [...] and found that they almost overlapped", which domain did you consider? As pointed out by @JohnHughes, this approximation becomes really bad as $x \to +\infty$ and it also becomes really bad when $x \to 0$ ($x > 0$). – pitchounet Jun 13 '19 at 15:03
  • Sry but , i just plotted the two graphs in desmos graph calculator. And was not able to distinguish the two . – kurama Jun 13 '19 at 15:20
  • @John Hughes use calculator and you will get it almost close – kurama Jun 13 '19 at 15:23
  • You still haven't said over what range of $x$ you did your graph. – lurker Jun 13 '19 at 17:59
  • I don't need to use a calculator -- I can do math myself. :) And doing it shows that the approximation is bad for large $x$, as expected. It's no surprise (See Ross's answer) that you can make it a good approximation for a modest range of $x$-values -- after all, $\log$ does have a Taylor series at $1$. – John Hughes Jun 13 '19 at 20:18
  • @lurker range was of the positive real numbers – kurama Jun 14 '19 at 03:03
  • That's the mathematical domain of course, but if you graphed it clearly you had a max $x$ for the graph. Your calculator isn't infinitely wide. ;) – lurker Jun 14 '19 at 03:08
  • @Mathlover Let's try $x=10^{10^{10}}$. Then $\log_{10}x=10^{10}$ but $x^{1/2^{19}}=10^{10^{10}/2^{19}}>10^{10^{10}/10^6}=10^{10^4}$, so $227695(x^{1/2^{19}}-1)>10^{10005}$ is way bigger than $10^{10}$. – user10354138 Jun 14 '19 at 04:07

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The approximation is of the form $\log_{10}x \approx a(x^b-1)$ for $b \ll 1$. This form assures that the value at $x=1$ is $0$, which agrees with the logarithm. The derivative is $abx^{b-1}$ while the derivative of $\log_{10}x$ is $\frac 1{x\log 10}$ When $b \ll 1$ and $x$ is not too large $x^b$ will be very close to $1$. We note that $$227695 \cdot 2^{-19}-\frac 1{\log 10} \approx 10^{-7}$$ so the derivatives are very close, which will make the curves very close until the $x^{2^{-19}}$ factor starts to matter. The approximation will be very good for any tiny $b$ when $a$ is chosen to make $ab \approx \frac 1{\log 10}$ The smaller $b$ gets the longer it will stay close. Once $x$ gets large enough the approximation will be terrible.

Ross Millikan
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  • In particular, when $x = 10^{1/b}$, (and when you chose $ab \approx \frac{1}{\log 10 }$ as Ross shows you must to get a good approximation near $x = 1$) you'll get the approximation $\log_{10} x \approx 9a \approx \frac{9}{b \log 10} \approx \frac{3.9}{b}$, where the correct answer is $1/b$, an error of almost $300%$. – John Hughes Jun 13 '19 at 20:32
  • Yeah , i got it . But then also the approximation is quite good for a large range of number , which can be used in our daily life calculations – kurama Jun 14 '19 at 03:06