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# The most effective method to compute natural Log?in model - Model

The most effective method to compute natural Log?in model - Model

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There are Three techniques for figuring of estimation of Natural Log :- (a) Using Natural log table (b) Using ex/e-x table (c) Using standard log tables I Method: Using Natural log tables Example 1 : Find the common log of (a) 0.75 (b) 1.24 (c) 1.0379 (d) 1.2667 Answer (a) ln(0.75) = - 0.2877 (b) ln(1.24) = 0.2151 (c) ln(1.0379) =? ln(1.03) = 0.0296 ln(1.04) = 0.0392 For LHS diff. of 0.01, RHS diff. is 0.0096 For LHS diff. of 1, RHS diff. is 0.96 For LHS diff. of 0.0079, RHS diff. is: (0.96)x 0.0079 i.e. 0.0076 ln (1.0379) = 0.0296 +0.0076 = 0.03702 (d) ln(1.2667) =? ln(1.26) = 0.2311 ln(1.27) = 0.2390 For LHS diff. of 0.01, RHS diff. is 0.0079 For LHS diff. of 1, RHS diff. is 0.79 For LHS diff. of 0.0067, RHS diff. is : (0.79)x 0.0067 i.e. 0.005293 ln (1.2667) = 0.2311 + 0.005293 = 0.236393

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II Method: Using ex/e-x table Let ex = y then, ln(y) =x Example 2 : Find common log of (i) 0.2567(ii) 0.3570 (iii)0.2466 (iv) 0.1979 and (v) 4.0960 Answers (i) Consulting the table of estimations of ex/e-x, we find : 0.2567 = e-1.36 Hence, characteristic log of 0.2567 = - 1.36 (ii) Consulting the table of estimations of ex/e-x, we find : 0.3570 = e-1.03 Hence, normal log of 0.3570 = - 1.03 (iii) Consulting the table of estimations of ex/e-x, we find : 0.2466 = e-1.40 Hence, regular log of 0.2466 = - 1.40 (iv) Consulting the table of estimations of ex/e-x, we find : 0.1979 = e-1.62 Hence, normal log of 0.1979 = - 1.62 (v) Consulting the table of estimations of ex/e-x, we find : 4.0960 = e1.41 Hence, normal log of 4.0960 = 1.41 Example 3 : Find the normal log of (a) 0.75 (b) 1.24 utilizing ex/e-x table Answer (a) ln(0.75) = ? e-0.28 = 0.7558 e-0.29 = 0.7483 0.7558 = e-0.28 0.7483 = e-0.29 For 0.0075 ? in LHS, e's energy ? by 0.01 For 1? in LHS, e's energy ? by 0.01/0.0075 For 0.0058 ? in LHS, e's energy ? by (0.01/0.0075)x0.0058 i.e. by 0.0077 = 0.7558 โ 0.0058 = e-0.28-0.0077 0.75 = e-0.287686 ln 0.75 = - 0.287686 (b) ln(1.24) = ? e0.21 = 1.2337 e0.22 = 1.2461 1.2337 = e0.21 1.2461 = e0.22 For 0.0124 ? in LHS, e's energy ? by 0.01 For 1 ? in RHS, e's energy ? by 0.01/0.0124 For 0.0063 ? in RHS, e's energy ? by (0.01/0.0124)x0.0063 i.e. 0.0050806 1.2337 +0.0063 = e0.21+0.0050806 = 1.24 = e0.215097 ln 1.24 = 0.215097

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III METHOD :- USING ORDINARY LOG TABLES ln (x) i.e. characteristic log = Ordinary log of x/0.4343 Example 4: Try Example 1 utilizing customary log tables Answer (a) : ln(0.75) = ? Log 0.75 = - 0.1249 ln 0.75 = - 0.1249/0.4343 = - 0.2876 (b) ln(1.24) = ? Log 1.24 = 0.0934 ln 1.24 = 0.0934/.4343 = 0.2151 (c) ln(1.0379) =? Log 1.0379 = Log 1.038 = 0.0162 ln 1.0379 = 0.0162/0.4343 = 0.0373 (d) ln(1.2667) = ? Log 1.2667 = Log 1.267 = 0.1028 ln 1.2667 = 0.1028/0.4343 = 0.2367

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