{"id":2050,"date":"2023-01-11T11:52:12","date_gmt":"2023-01-11T10:52:12","guid":{"rendered":"https:\/\/mxth.dk\/?page_id=2050"},"modified":"2023-01-31T10:27:35","modified_gmt":"2023-01-31T09:27:35","slug":"opgaver-til-logaritme","status":"publish","type":"page","link":"https:\/\/mxth.dk\/?page_id=2050","title":{"rendered":"Opgaver til logaritme"},"content":{"rendered":"\n<p class=\" eplus-wrapper\">En lille samling af ekstraopgaver til emnet logritmefiunktioner<\/p>\n\n\n<p class=\" eplus-wrapper eplus-styles-uid-a61566\"><strong>Logaritmefunktioner &#8211; en introduktion<\/strong><\/p>\n\n\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_77261c-8b kt-accordion-has-7-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"false\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_0027f0-a1\" id=\"LEI001\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI001<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">L\u00f8s f\u00f8lgende udtryk<\/p>\n\n\n<ul class=\"eplus-wrapper wp-block-list eplus-styles-uid-23ae66\">\n<li class=\" eplus-wrapper\">$27=\\frac{12}{1-\\frac{1}{2}\\cdot e^{-x}}$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$1000 = \\frac{10000}{1+19\\cdot e^{-t}}$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$300 = \\frac{400}{1+3\\cdot e^{-2k}}$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$16^x+4^x-6=0$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$\\frac{e^x+e^{-x}}{e^x-e^{-x}}=6$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$\\frac{ln(4\\cdot x+2)}{ln(4\\cdot x-2)}=2$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$\\frac{e^x-e^{-x}}{e^x+e^{-x}}=\\frac{1}{2}$<\/li>\n<\/ul><\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-3 kt-pane_5428da-fc\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI002<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Et spektrofotometer m\u00e5ler koncentrationen af et stof opl\u00f8st i vand ved at sende lys igennem opl\u00f8sningen og m\u00e5le hvor meget af lyset der kommer igennem. Med andre ord, hvis vi kender hvor meget lys der er absorberet s\u00e5 kan vi kan vi beregne koncentrationen af vores pr\u00f8ve. For et givent stof er koncentrationen (i mol pr liter) fundet ved at bruge f\u00f8lgende formel<\/p>\n\n\n\n<p class=\"has-text-align-center eplus-wrapper\">$C=-2500\\cdot ln(\\frac{I}{I_0})$<\/p>\n\n\n\n<p class=\" eplus-wrapper\">hvor $I_0$ er intensiteten af det lys pr\u00f8ven belyses med og $I$ er intensiteten af der lys som kommer igennem pr\u00f8ven. Find koncentrationen for stoffet hvis intensiteten $I$ er 70% af $I_0$.<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized eplus-wrapper\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/mxth.dk\/wp-content\/uploads\/2023\/01\/image.png?resize=406%2C147&#038;ssl=1\" alt=\"\" class=\"wp-image-2071\" width=\"406\" height=\"147\"\/><\/figure>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-4 kt-pane_6cff97-ff\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI003<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Alderen p\u00e5 en gammel artefakt kan bestemmes ud fra m\u00e6ngden af radioaktivt kulstof-14, der er tilbage i den. Hvis $D_0$ er den oprindelige m\u00e6ngde kulstof-14 og $D$ er den resterende m\u00e6ngde, s\u00e5 er artifaktens alder (i \u00e5r) givet ved<\/p>\n\n\n\n<p class=\"has-text-align-center eplus-wrapper\">$A=-8267\\cdot ln(\\frac{D}{D_0})$<\/p>\n\n\n\n<p class=\" eplus-wrapper\">Find alderen p\u00e5 et objekt, hvis m\u00e6ngde af $D$ af kulstof-14, der er tilbage i objektet, er 73% af den oprindelige m\u00e6ngde $D_0$.<\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-5 kt-pane_24c659-40\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI004<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">En bestem bakteriestamme deler sig hver tredje time. Hvis en koloni startes med 50 bakterier, s\u00e5 er tiden t (i timer), der kr\u00e6ves for at kolonien vokser til N bakterier, givet ved <\/p>\n\n\n\n<p class=\"has-text-align-center eplus-wrapper\">$t=3\\cdot \\frac{log(N\/50)}{log(2)}$<\/p>\n\n\n\n<p class=\" eplus-wrapper\">Find den tid, det tager for kolonien at vokse til en million bakterier.<\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-6 kt-pane_77350e-19\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI005<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Hvilket udtryk er st\u00f8rst. $log_4(17)$ eller $log_5(24)$? Begrund dit svar.<\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-7 kt-pane_087215-fc\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LEI006<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Sammenlig $log_{10}(1000)$ til antallet af cifre i 1000. G\u00f8r det samme for 10.000. Hvor mange cifre har et hvilket som helst heltal mellem 1000 og 10.000? Mellem hvilke v\u00e6rdier vil 10-tals-logaritmen til s\u00e5danne et tal ligge? Brug denne observation til at forklar hvorfor antallet af cifre i et vilk\u00e5rligt positivt heltal x er $\\lfloor log_{10}(x) \\rfloor +1$. Hvor mange cifre har tallet $2^{100}$?<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n<p class=\" eplus-wrapper eplus-styles-uid-e8ada0\"><strong>Logaritmeregneregler<\/strong><\/p>\n\n\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_8973ee-29 kt-accordion-has-3-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"false\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_7c2423-63\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LRR001<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Udvid hver af disse udtryk<\/p>\n\n\n<ol class=\"eplus-wrapper wp-block-list eplus-styles-uid-3d8100\">\n<li class=\" eplus-wrapper\">$log_9(6x^3y^5z)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$log_3(\\frac{p^2q}{\\sqrt[5]{3q-1}})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$log_{11}(ab^{-4}c^12d^7)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$log_4(10t^2uv^{-3})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(\\frac{x^7}{\\sqrt[3]{x+2}}$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$log_7(h^2j^{11}k^{-5})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">\n\n\n\n<li class=\" eplus-wrapper\">$log_5(a^6b^{-3}c^4)$<\/li>\n<\/ol><\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-2 kt-pane_172a80-10\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LRR002<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Reducer hvert udtryk<\/p>\n\n\n<ol class=\"eplus-wrapper wp-block-list eplus-styles-uid-91b9af\">\n<li class=\" eplus-wrapper\">$3\\cdot log_5(x)-\\frac{1}{2}\\cdot log_5(6-x)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$5\\cdot log_7(2x)-\\frac{1}{3}\\cdot log_7(5x+1)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$7\\cdot log_3(a)+log_3(b)-2\\cdot log_3(8c)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$4\\cdot ln(x+3)-\\frac{1}{5}\\cdot ln(4x+7)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$2\\cdot log_8(9x)-log_8(2x-5)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(13)+7\\cdot ln(a)-11\\cdot ln(b) + ln(c)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$2\\cdot log_6(5a)+log_6(b)+7\\cdot log_6(c)$<\/li>\n<\/ol><\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-3 kt-pane_b6d5ea-ec\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><span class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">LRR003<\/span><\/span><span class=\"kt-blocks-accordion-icon-trigger\"><\/span><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p class=\" eplus-wrapper\">Udtryk hvert af f\u00f8lgende udtryk i form af $log_2$ og $log_5$<\/p>\n\n\n<ol class=\"eplus-wrapper wp-block-list eplus-styles-uid-e85b38\">\n<li class=\" eplus-wrapper\">$ln(\\frac{4}{5})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(80)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(\\frac{0,8}{2})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(2000)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(200)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(12,5)$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(\\frac{2}{5})$<\/li>\n\n\n\n<li class=\" eplus-wrapper\">$ln(1,6)$<\/li>\n<\/ol><\/div><\/div><\/div>\n<\/div><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>En lille samling af ekstraopgaver til emnet logritmefiunktioner Logaritmefunktioner &#8211; en introduktion Logaritmeregneregler<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"ub_ctt_via":"","editor_plus_copied_stylings":"{}","footnotes":""},"categories":[3,4],"tags":[],"class_list":["post-2050","page","type-page","status-publish","hentry","category-matematik","category-opgave"],"featured_image_src":null,"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/pages\/2050","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2050"}],"version-history":[{"count":16,"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/pages\/2050\/revisions"}],"predecessor-version":[{"id":2079,"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/pages\/2050\/revisions\/2079"}],"wp:attachment":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2050"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2050"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2050"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}