{"id":63,"date":"2022-01-27T17:58:00","date_gmt":"2022-01-27T16:58:00","guid":{"rendered":""},"modified":"2023-08-04T22:45:52","modified_gmt":"2023-08-04T20:45:52","slug":"logaritmicke-tabulky","status":"publish","type":"post","link":"https:\/\/www.jankafialka.sk\/?p=63","title":{"rendered":"Logaritmick\u00e9 tabu\u013eky"},"content":{"rendered":"\n

Ob\u010das, pri prehrab\u00e1van\u00ed sa v star\u00fdch knih\u00e1ch, mo\u017eno narazi\u0165 na podivn\u00fa knihu pln\u00fa \u010d\u00edsel. Logaritmick\u00e9 tabu\u013eky. Kedysi bola takmer v ka\u017edej dom\u00e1cnosti. Dnes u\u017e m\u00e1lokto vie, na \u010do je v\u00f4bec dobr\u00e1. Tie\u017e som tak\u00fa objavila v kni\u017enici po mojom dedu\u0161kovi. Je to tak\u00fd artefakt z d\u00e1vnych \u010dias. Z \u010dias pred internetom, pred kalkula\u010dkami a pred po\u010d\u00edta\u010dmi. \u00c1no, na\u0161i predkovia dok\u00e1zali \u017ei\u0165 aj bez t\u00fdchto vec\u00ed. A dok\u00e1zali po\u010d\u00edta\u0165. Nielen plus, m\u00ednus, kr\u00e1t, \u010di deleno. Vypo\u010d\u00edtali napr\u00edklad aj odmocninu s takou presnos\u0165ou, ak\u00fa n\u00e1m dnes d\u00e1 kalkula\u010dka.<\/p>\n\n\n\n

\"\"
Kni\u017enica<\/figcaption><\/figure>\n\n\n\n

Pou\u017eili pri tom pr\u00e1ve tak\u00fato k\u00fazeln\u00fa knihu. Napr\u00edklad druh\u00fa odmocninu z dvoch vypo\u010d\u00edtali tak, \u017ee v tabu\u013ek\u00e1ch na\u0161li logaritmus dvoch, v\u00fdsledok vydelili dvoma a zasa v tabu\u013ek\u00e1ch na\u0161li, ko\u013eko je desa\u0165 umocnen\u00fdch na toto \u010d\u00edslo. Pre t\u00fdch, ktor\u00ed vedia, ako funguje funkcia logaritmus so z\u00e1kladom desa\u0165, nie je ve\u013ek\u00fd probl\u00e9m roz\u0161ifrova\u0165 fungovanie logaritmick\u00fdch tabuliek.<\/p>\n\n\n\n

Logaritmus<\/h2>\n\n\n\n

Po\u010fme si ale tro\u0161ku pripomen\u00fa\u0165, \u010do je to vlastne ten logaritmus. (V celom tomto pr\u00edspevku sa bav\u00edm len s dekadick\u00fdm logaritmom.) Jednoducho povedan\u00e9, je to \u010d\u00edslo, na ktor\u00e9 mus\u00edm umocni\u0165 desiatku, aby som dostala logaritmovan\u00e9 \u010d\u00edslo. Napr\u00edklad logaritmus z desiatky je jeden, preto\u017ee desiatku mus\u00edm umocni\u0165 na prv\u00fa, aby som dostala v\u00fdsledok desa\u0165. Logaritmus z jednotky je nula, preto\u017ee desa\u0165 na nult\u00fa je jeden. No, a toto s\u00fa hrani\u010dn\u00e9 hodnoty, pre ktor\u00e9 sta\u010d\u00ed ma\u0165 v\u00fdsledky v logaritmick\u00fdch tabu\u013ek\u00e1ch. \u010ci\u017ee, \u010d\u00edsla, ktor\u00e9 logaritmujeme, id\u00fa od jednotky po desiatku a v\u00fdsledky od nuly po jednotku. Jednotliv\u00e9 tabu\u013eky sa l\u00ed\u0161ia len v presnosti: na ko\u013eko desatinn\u00fdch miest, resp. platn\u00fdch \u010d\u00edslic, po\u010d\u00edtame.<\/p>\n\n\n\n

Na obr\u00e1zku ni\u017e\u0161ie je uk\u00e1\u017eka z najjednoduch\u0161\u00edch tabuliek, ktor\u00e9 s\u00fa s\u00fa\u010das\u0165ou matematicko-fyzik\u00e1lno-chemick\u00fdch tabuliek. Logaritmick\u00e9 tabu\u013eky v nich zaberaj\u00fa len nieko\u013eko m\u00e1lo str\u00e1nok a po\u010d\u00edtaj\u00fa s presnos\u0165ou na 5 miest. Desatinn\u00e9 \u010diarky ignorujeme, ke\u010f\u017ee vieme, v akom rozp\u00e4t\u00ed sa pohybujeme. (\u010c\u00edsla, ktor\u00e9 logaritmujeme s\u00fa medzi 1 a 10 a v\u00fdsledky medzi nulou a jednotkou.) Po\u010fme sa pozrie\u0165 na t\u00fa odmocninu z dvoch. Logaritmus dvojky n\u00e1jdeme v riadku 200 – to s\u00fa prv\u00e9 tri miesta, v st\u013apci 0- \u0161tvrt\u00e9 miesto. Logaritmus dvojky je 0,30103. Pre\u010do teraz budem po\u010d\u00edta\u0165 polovicu? Logaritmus m\u00e1 jednu \u00fa\u017easn\u00fa vlastnos\u0165: ak \u010d\u00edslo, ktor\u00e9 logaritmujeme m\u00e1 nejak\u00fa mocninu, t\u00fa m\u00f4\u017eeme vybra\u0165 pred logaritmus a vyn\u00e1sobi\u0165. T.j. log2^{1\/2}= 1\/2\u2022log2. (Zabudla som e\u0161te podotkn\u00fa\u0165, \u017ee druh\u00e1 odmocnina je to ist\u00e9 ako mocnina s exponentom 1\/2.)<\/p>\n\n\n\n

Pokra\u010dujem: polovica z 0,30103 je 0,150515 a toto \u010d\u00edslo mus\u00edm n\u00e1js\u0165 vo v\u00fdsledkoch. Na\u0161\u0165astie, logaritmus je prost\u00e1 funkcia – s rast\u00facim x<\/i> rastie aj logaritmus, tak\u017ee v\u00fdsledky id\u00fa pekne po poriadku. Najbli\u017e\u0161ie vid\u00edm 15045 (hviezdi\u010dka pred 045 znamen\u00e1, \u017ee aj ke\u010f je v riadku so za\u010diatkom 14, u\u017e m\u00e1 za\u010diatok \u010fal\u0161ieho riadku, t.j. 15. Tak\u017ee najbli\u017e\u0161\u00ed v\u00fdsledok by bol 1,414. To je presnos\u0165 4 miesta. M\u00f4\u017eeme to e\u0161te k\u00fasok upresni\u0165 pomocou tabule\u010dky vpravo. K 15045 by sme mali prida\u0165 asi \u0161estku, aby sme boli bli\u017e\u0161ie k 150515. Na konci n\u00e1\u0161ho riadku je pozn\u00e1mka 31 a v st\u013ap\u010deku s ozna\u010den\u00edm 31 v tabule\u010dke ved\u013ea je celkom fajn hodnota 6,2. To je skoro presne to\u013eko, ko\u013eko sme chceli. Riadok 2, tak\u017ee piate miesto je 2. Odmocnina z dvojky je teda 1,4142 s presnos\u0165ou na p\u00e4\u0165 miest.<\/p>\n\n\n\n

\"Logaritmick\u00e9
Logaritmick\u00e9 tabu\u013eky<\/figcaption><\/figure>\n\n\n\n

Rovnica, ktor\u00e1 opisuje tento postup by vyzerala asi takto:<\/p>\n\n\n\n

2^{1\/2} = x<\/i><\/p>\n\n\n\n

log 2^{1\/2} = log x<\/i><\/p>\n\n\n\n

1\/2\u2022 log2 = log x<\/i><\/p>\n\n\n\n

1\/2\u2022 0,30103 = log x<\/p>\n\n\n\n

0,150515 = log x<\/p>\n\n\n\n

x<\/i> = 1,4142<\/p>\n\n\n\n

(Namiesto rovn\u00edtok by tam mali by\u0165 rovn\u00edtka s bodkou, ako pribli\u017ene, ale p\u00ed\u0161em z mobilu, tak\u017ee mi to, h\u00e1dam, odpust\u00edte.)<\/p>\n\n\n\n

Ako je ale mo\u017en\u00e9, \u017ee mi sta\u010d\u00ed vedie\u0165 logaritmy jednocifern\u00fdch \u010d\u00edsel? S t\u00fdm s\u00favis\u00ed \u010fal\u0161ia kr\u00e1sna vlastnos\u0165 logaritmov. Logaritmus v\u00e4\u010d\u0161ieho (alebo aj men\u0161ieho) \u010d\u00edsla \u013eahko odvod\u00edme. Napr\u00edklad logaritmus \u010d\u00edsla 250 je logaritmus s\u00fa\u010dinu 2,5\u2022100 a logaritmus<\/b> s\u00fa\u010dinu<\/b> je s\u00fa\u010det logaritmov<\/b>. T.j. sta\u010d\u00ed ur\u010di\u0165 logaritmus 2,5 a pripo\u010d\u00edta\u0165 k nemu logaritmus stovky, \u010do je dvojka. (Desa\u0165 na ko\u013ek\u00fa je sto?) Opa\u010dne, ke\u010f mi vyjde, \u017ee h\u013eadan\u00e9 \u010d\u00edslo m\u00e1 logaritmus 3,78, sta\u010d\u00ed n\u00e1js\u0165 \u010d\u00edslo prisl\u00fachaj\u00face v\u00fdsledku 0,78 a toto vyn\u00e1sobi\u0165 tis\u00edckou. (Desa\u0165 na 3+0,78 je desa\u0165 na tretiu kr\u00e1t desa\u0165 na 0,78.)<\/p>\n\n\n\n

Vyu\u017eitie<\/h2>\n\n\n\n

S logaritmick\u00fdmi tabu\u013ekami dok\u00e1\u017eeme teda po\u010d\u00edta\u0165 ak\u00e9ko\u013evek mocniny a odmocniny z ak\u00e9hoko\u013evek \u010d\u00edsla. Logaritmick\u00e9 tabu\u013eky m\u00f4\u017eeme pou\u017ei\u0165 aj na r\u00fdchlej\u0161ie n\u00e1sobenie. Pozrieme logaritmy oboch \u010dinite\u013eov. S\u010d\u00edtame ich a n\u00e1jdeme \u010d\u00edslo, ktor\u00e9mu prisl\u00facha v\u00fdsledok. Pou\u017e\u00edvame tak znova vlastnos\u0165: logaritmus s\u00fa\u010dinu je s\u00fa\u010det logaritmov.<\/p>\n\n\n\n

Mo\u017eno sa v\u00e1m cel\u00e9 toto nezd\u00e1 v\u00f4bec nejak\u00e9 \u00fa\u017easn\u00e9. Zamyslite sa ale nad t\u00fdm, ako tieto tabu\u013eky vznikli. Je sedemn\u00e1ste storo\u010die. \u017diadna kalkula\u010dka neexistuje. Po\u010d\u00edta\u0165 odmocninu z dvojky tak, \u017ee sk\u00fa\u0161ate r\u00f4zne \u010d\u00edsla, ktor\u00e9 n\u00e1sob\u00edte sam\u00e9 sebou, a\u017e k\u00fdm v\u00e1m nevyjde takmer presne dvojka, je dos\u0165 zd\u013ahav\u00e9 a pri v\u00fdpo\u010dte inej odmocniny v\u00e1m v\u00f4bec nepom\u00f4\u017ee, \u017ee u\u017e ste po\u010d\u00edtali podobn\u00fd pr\u00edklad. Potom sa n\u00e1jde niekto, kto je ochotn\u00fd venova\u0165 neskuto\u010dne ve\u013ea \u010dasu na mraven\u010diu pr\u00e1cu: vypo\u010d\u00edta logaritmy v\u0161etk\u00fdch \u010d\u00edsel od jeden do desa\u0165 id\u00fac po stotin\u00e1ch, nesk\u00f4r tis\u00edcin\u00e1ch aj desa\u0165tis\u00edcin\u00e1ch. Sta\u010d\u00ed to spravi\u0165 raz, zap\u00edsa\u0165 do knihy a u\u013eah\u010d\u00ed t\u00fdm v\u00fdpo\u010dty v\u0161etk\u00fdm po \u0148om.<\/p>\n\n\n\n

Ako to v\u00f4bec bez kalkula\u010dky mohol zvl\u00e1dnu\u0165? Odpove\u010f nepozn\u00e1m, ale ak by som to mala robi\u0165 ja, postupovala by som asi takto. Vytv\u00e1rala by som opa\u010dn\u00fa tabu\u013eku: tabu\u013eku mocn\u00edn desiatky a ke\u010f by bola hotov\u00e1, len by som zap\u00edsala v\u00fdsledky ako n<\/i> a vstupn\u00e9 \u010d\u00edsla ako v\u00fdsledky. Za\u010dala by som s desatinami, tak\u017ee prv\u00e9 by som musela ur\u010di\u0165, ko\u013eko je desa\u0165 na 0,1, \u010di\u017ee desiatu odmocninu z desa\u0165. No, toto by trvalo dlho, ke\u010f\u017ee by som to proste sk\u00fa\u0161ala: tipnem \u010d\u00edslo, desa\u0165kr\u00e1t ho vyn\u00e1sob\u00edm sam\u00e9 sebou a pozriem, \u010di som dos\u0165 bl\u00edzko k desiatke. S uspokojiv\u00fdm v\u00fdsledkom \u013eahko dorob\u00edm aj mocniny 0,2, 0,3, 0,4, … , 0,9, ke\u010f\u017ee 0,2 je 0,1+0,1 a to v mocnine znamen\u00e1, \u017ee sta\u010d\u00ed da\u0165 desa\u0165 na 0,1 kr\u00e1t zasa desa\u0165 na 0,1 a m\u00e1m v\u00fdsledok. Tento zasa m\u00f4\u017eem vyn\u00e1sobi\u0165 s desa\u0165 na 0,1 a budem ma\u0165 desa\u0165 na 0,3 a t.\u010f. To by som vlastne mala hotov\u00e9 u\u017e vtedy, ke\u010f som tipovala a zis\u0165ovala, \u010di mi vyjde desa\u0165.<\/p>\n\n\n\n

Desa\u0165 na 0,01 by som zasa musela tipova\u0165, ale, samozrejme, nebudem po\u010d\u00edta\u0165 st\u00fa mocninu a \u010daka\u0165, kedy vyjde desa\u0165. Sta\u010d\u00ed desiatu a mus\u00ed vyjs\u0165 to, \u010do m\u00e1m u\u017e ur\u010den\u00e9 – t\u00fdch desa\u0165 na 0,1. (Lebo desa\u0165 na 0,01 je to ist\u00e9 ako desa\u0165 na 0,1 a to cel\u00e9 e\u0161te na 0,1.) Ostatn\u00e9 \u010d\u00edsla zasa dopo\u010d\u00edtam \u013eahko. Napr\u00edklad desa\u0165 na 0,22 je desa\u0165 na 0,2+0,01+0,01, \u010do je s\u00fa\u010din desa\u0165 na 0,2 kr\u00e1t desa\u0165 na 0,01 kr\u00e1t desa\u0165 na 0,01. Takto nejako si viem predstavi\u0165, \u017ee by sa logaritmick\u00e9 tabu\u013eky dali vytvori\u0165. Ale robi\u0165 by sa mi to, popravde, nechcelo. \u013dudia, ktor\u00ed ich naozaj spravili, si zasl\u00fa\u017eia n\u00e1\u0161 obdiv a \u00factu. Prv\u00fdm z nich bol Henry<\/b> Briggs<\/b>. Jemu aj pripisujeme objav dekadick\u00e9ho logaritmu. Nadviazal na pr\u00e1cu Johna<\/b> Napiera<\/b>, ktor\u00fd vyna\u0161iel logaritmus, ale pri z\u00e1klade e:<\/i> prirodzen\u00fd logaritmus. \ud83d\udc4f<\/p>\n\n\n\n

Ke\u010f teda niekde natraf\u00edte na kni\u017eku pln\u00fa \u010d\u00edsel s n\u00e1zvom Logaritmick\u00e9 tabu\u013eky sedemmiestne alebo aj viacmiestne, nevyhadzujte ju. Niekto venoval ve\u013ea \u010dasu a \u00fasilia pri jej tvorbe. A navy\u0161e, m\u00f4\u017ee sa v\u00e1m z\u00eds\u0165, ke\u010f nebude \u017eiadna kalkula\u010dka poruke.<\/p>\n","protected":false},"excerpt":{"rendered":"

Ob\u010das, pri prehrab\u00e1van\u00ed sa v star\u00fdch knih\u00e1ch, mo\u017eno narazi\u0165 na podivn\u00fa knihu pln\u00fa \u010d\u00edsel. Logaritmick\u00e9 tabu\u013eky. Kedysi bola takmer v ka\u017edej dom\u00e1cnosti. Dnes u\u017e m\u00e1lokto vie, na \u010do je v\u00f4bec dobr\u00e1. Tie\u017e som tak\u00fa objavila v kni\u017enici po mojom dedu\u0161kovi. Je to tak\u00fd artefakt z d\u00e1vnych \u010dias. Z \u010dias pred internetom, pred kalkula\u010dkami a pred […]<\/p>\n","protected":false},"author":2,"featured_media":1942,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[17,18,19],"class_list":["post-63","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nezaradene","tag-historia","tag-logaritmus","tag-mocnina"],"yoast_head":"\nLogaritmick\u00e9 tabu\u013eky - jankafialka<\/title>\n<meta name=\"description\" content=\"Logaritmick\u00e9 tabu\u013eky s\u00fa pre mnoh\u00fdch u\u017e zastaranou z\u00e1le\u017eitos\u0165ou. 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