Natural Logs
Key Questions
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The Answer is 1.
ln(e)ln(e) is the same thing aslog_e(e)loge(e) Because
e^1 = e, log_e(e) = 1e1=e,loge(e)=1 
Kevin B.
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1
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Mar 24 2015
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Answer:
You can approximate
ln xlnx by approximatingint_1^x 1/t dt∫x11tdt using Riemann sums with the trapezoidal rule or better with Simpson's rule.Explanation:
For example, to approximate
ln(7)ln(7) , split the interval[1, 7][1,7] into a number of strips of equal width, and sum the areas of the trapezoids with vertices:(x_n, 0)(xn,0) ,(x_n, 1/x_n)(xn,1xn) ,(x_(n+1), 0)(xn+1,0) ,(x_(n+1), 1/(x_(n+1)))(xn+1,1xn+1) If we use strips of width
11 , then we get six trapezoids with average heights:3/434 ,5/12512 ,7/24724 ,9/40940 ,11/601160 ,13/841384 .If you add these, you find:
3/4+5/12+7/24+9/40+11/60+13/84 ~~ 2.0234+512+724+940+1160+1384≈2.02 If we use strips of width
1/212 , then we get twelve trapezoids with average heights:5/656 ,7/12712 ,9/20920 ,...,25/15625156 ,27/18227182 Then the total area is:
1/2 xx (5/6+7/12+9/20+...+25/156+27/182) ~~ 1.97 Actually
ln(7) ~~ 1.94591 , so these are not particularly accurate approximations.Simpson's rule approximates the area under a curve using a quadratic approximation. For a given
h > 0 , the area under the curve off(x) betweenx_0 andx_0 + 2h is given by:h/3(f(x_0) + 4f(x_0+h) + f(x_0+2h)) Try improving the approximation using Simpson's rule, using
h = 1/2 then we get six approximate areas to sum:h/3(25/6+73/30+145/84+241/180+361/330+505/546)~~1.947 That's somewhat better.
George C.
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3
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Nov 4 2015
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ln(x) is the same as sayinglog_e(x) .e is the base of the natural log.
Kevin B.
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1
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Apr 5 2015