N\u00e5r vi skal regne med logaritme er der nogle regneregler som er rare at kende. De er som f\u00f8lger<\/p>\n
$$\\log(a\\cdot b) = \\log(a) + \\log(b)$$
\n$$\\log (\\dfrac{a}{b}) = \\log(a)-\\log(b)$$
\n$$\\log(a^n) = n\\cdot\\log(a)$$<\/p>\n
Vi kan se at logaritmen ikke har nogen base, og det er fordi at disse regneregler g\u00e6lder uanset basen p\u00e5 logaritmen.<\/p>\n
Hvis vi skal forklare hvorfor regnereglerne er som de er skal vi benytte potensregnereglen<\/p>\n
$$a^n \\cdot a^m = a^{n\\cdot m}$$<\/p>\n
samt at logaritme og s\u00e6tte i potens udligner hinanden hvis basen og grundtallet er ens<\/p>\n
$$\\log_{10} (10^x)=x$$
\n$$10^{\\log_{10}(x)}=x$$<\/p>\n
Vi kan nu forklare den f\u00f8rste logaritmeregneregel<\/p>\n
$$\\log_{10}(a\\cdot b)=\\log_{10}(10^{\\log_{10}(a)}\\cdot 10^{\\log_{10}(b)}) \\\\= \\log_{10}(10^{\\log_{10}(a)+\\log_{10}(b)})=\\log_{10}(a)+\\log_{10}(b)$$<\/p>\n
Vi kan forklare den anden logaritmeregneregel p\u00e5 tilsvarende m\u00e5de<\/p>\n
$$\\log_{10}(\\dfrac{a}{b})=\\log_{10}(\\dfrac{10^{\\log_{10}(a)}}{10^{\\log_{10}(b)}})\\\\ =\\log_{10}(10^{\\log_{10}(a)-\\log_{10}(b)}) = \\log_{10}(a)-\\log_{10}(b)$$<\/p>\n
Hvis vi skal forklare den sidste kan vi benytte den f\u00f8rste logaritme regneregel<\/p>\n
$$\\log_{10}(a^n)=\\log_{10}(\\underbrace{a\\cdot\\ldots\\cdot a}_{\\text{n gange}})=\\underbrace{\\log_{10}(a)\\cdot\\ldots\\cdot\\log_{10}(a)}_{\\text{n gange}}=n\\cdot\\log_{10}(a)$$<\/p>\n
Til at forklare disse logaritmeregneregler er der benyttet log-10, men man kunne have gjort tilsvarende for andre logaritmer med en vilk\u00e5rlig base.<\/p>\n
Vi har derfor at logaritmeregnereglerne er som f\u00f8lger<\/p>\n
$$\\log_{b}(a\\cdot b) = \\log_{b}(a) + \\log_{b}(b)$$
\n$$\\log_{b}(\\dfrac{a}{b}) = \\log_{b}(a)-\\log_{b}(b)$$
\n$$\\log_{b}(a^n) = n\\cdot \\log_{b}(a)$$<\/p>\n
Vi kan nu ogs\u00e5 vise at det er underordnet hvilken logaritme vi benytter for at l\u00f8se en eksponentiel ligning. Lad os se p\u00e5 et eksempel<\/p>\n
Vi skal l\u00f8se ligningen<\/p>\n
$$3\\cdot 1,03^x=9$$<\/p>\n
Vi starter med at reducere<\/p>\n
$$1,03^x=\\dfrac{9}{3}=3$$<\/p>\n
her kunne vi tage logaritmen med basen $1,03$ men lad os tage titals-logaritmen<\/p>\n
$$\\log_{10}(1,03^x)=\\log_{10}(3)$$<\/p>\n
og her kan vi benytte vores tredje regneregel<\/p>\n
$$x\\cdot\\log_{10}(1,03)=\\log_{10}(3)$$
\n$$x=\\dfrac{\\log_{10}(3)}{\\log_{10}(1,03)}=\\dfrac{0,477121}{0,012837}=37,167$$<\/p>\n","protected":false},"excerpt":{"rendered":"
N\u00e5r vi skal regne med logaritme er der nogle regneregler som er rare at kende. De er som f\u00f8lger $$\\log(a\\cdot b) = \\log(a) + \\log(b)$$ $$\\log (\\dfrac{a}{b}) = \\log(a)-\\log(b)$$ $$\\log(a^n) = n\\cdot\\log(a)$$ Vi kan se at logaritmen ikke har nogen base, og det er fordi at disse regneregler g\u00e6lder uanset basen p\u00e5 logaritmen. Hvis vi […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ub_ctt_via":"","editor_plus_copied_stylings":"{}","_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[36,3],"tags":[67],"class_list":["post-4053","post","type-post","status-publish","format-standard","hentry","category-logaritmefunktioner","category-matematik","tag-hhx"],"featured_image_src":null,"author_info":{"display_name":"Henriksen","author_link":"https:\/\/mxth.dk\/?author=1"},"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/posts\/4053","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/types\/post"}],"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=4053"}],"version-history":[{"count":2,"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/posts\/4053\/revisions"}],"predecessor-version":[{"id":4056,"href":"https:\/\/mxth.dk\/index.php?rest_route=\/wp\/v2\/posts\/4053\/revisions\/4056"}],"wp:attachment":[{"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4053"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4053"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mxth.dk\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4053"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}