{"id":222,"date":"2026-03-05T11:42:35","date_gmt":"2026-03-05T10:42:35","guid":{"rendered":"https:\/\/zalapto.pl\/baza\/?p=222"},"modified":"2026-03-05T11:46:46","modified_gmt":"2026-03-05T10:46:46","slug":"zadanie-4-matura-probna-matematyka-marzec-2026","status":"publish","type":"post","link":"https:\/\/zalapto.pl\/baza\/zadanie-4-matura-probna-matematyka-marzec-2026\/","title":{"rendered":"Zadanie 4 \u2013 Matura pr\u00f3bna matematyka CKE (Marzec 2026)"},"content":{"rendered":"\n\n\n<div class=\"mb-4\">\n    <a href=\"https:\/\/zalapto.pl\/baza\/matura-probna-matematyka-cke-2026-marzec\" class=\"btn btn-outline-primary btn-sm rounded-pill fw-bold shadow-sm\">\n        <i class=\"bi bi-arrow-left me-1\"><\/i> Wr\u00f3\u0107 do pe\u0142nego arkusza z marcem 2026\n    <\/a>\n<\/div>\n\n<div class=\"card border border-light shadow-sm rounded-4 mb-5\">\n    <div class=\"card-body p-4 p-md-5\">\n        \n        <h2 class=\"fw-bold text-dark mb-4 border-bottom pb-2 fs-4\">Tre\u015b\u0107 zadania<\/h2>\n        <div class=\"fs-5 mb-4 text-dark\" style=\"line-height: 1.8;\">\n            Doko\u0144cz zdanie. Wybierz w\u0142a\u015bciw\u0105 odpowied\u017a spo\u015br\u00f3d podanych.<br><br>\n            Liczba \\(\\log_{8}\\sqrt[5]{2}\\) jest r\u00f3wna:<br>\n            A. \\(\\frac{1}{15}\\)<br>\n            B. \\(-\\frac{1}{15}\\)<br>\n            C. \\(\\frac{3}{5}\\)<br>\n            D. \\(\\frac{5}{3}\\)\n        <\/div>\n\n        <h2 class=\"fw-bold text-dark mb-4 border-bottom pb-2 mt-5 fs-4\">Rozwi\u0105zanie krok po kroku<\/h2>\n        \n        <details class=\"mb-3\">\n            <summary class=\"btn btn-warning text-dark fw-bold rounded-pill px-4\" style=\"cursor: pointer;\">\n                <i class=\"bi bi-lightbulb\"><\/i> Zobacz podpowied\u017a\n            <\/summary>\n            <div class=\"p-4 mt-3 bg-light border rounded-4\">\n                Aby obliczy\u0107 ten logarytm bez zgadywania, sprowad\u017a podstaw\u0119 logarytmu (\\(8\\)) i liczb\u0119 logarytmowan\u0105 (\\(\\sqrt[5]{2}\\)) do pot\u0119gi o tej samej podstawie, czyli liczby \\(2\\). Pami\u0119taj, \u017ce pierwiastek mo\u017cna zapisa\u0107 jako pot\u0119g\u0119 u\u0142amkow\u0105!\n            <\/div>\n        <\/details>\n\n        <details>\n            <summary class=\"btn btn-success fw-bold rounded-pill px-4\" style=\"cursor: pointer;\">\n                <i class=\"bi bi-check-circle\"><\/i> Sprawd\u017a pe\u0142ne rozwi\u0105zanie\n            <\/summary>\n            <div class=\"p-4 mt-3 bg-success bg-opacity-10 border border-success rounded-4 text-dark fs-6\">\n                \n                <p><strong>1. Zapisujemy liczby jako pot\u0119gi dw\u00f3jki:<\/strong><\/p>\n                Wiemy, \u017ce \\(8 = 2^3\\).<br>\n                Z w\u0142asno\u015bci pot\u0119g wiemy r\u00f3wnie\u017c, \u017ce pierwiastek stopnia pi\u0105tego mo\u017cemy zapisa\u0107 jako pot\u0119g\u0119 u\u0142amkow\u0105: \\(\\sqrt[5]{2} = 2^{\\frac{1}{5}}\\).\n                \n                <p><strong>2. Podstawiamy to do naszego logarytmu:<\/strong><\/p>\n                $$ \\log_{8}\\sqrt[5]{2} = \\log_{2^3}2^{\\frac{1}{5}} $$\n                \n                <p><strong>3. Korzystamy ze wzor\u00f3w na logarytmy:<\/strong><\/p>\n                Wyrzucamy wyk\u0142adniki pot\u0119g przed logarytm. Wyk\u0142adnik z liczby logarytmowanej &#8222;wypada&#8221; do licznika, a wyk\u0142adnik z podstawy logarytmu do mianownika:\n                <div class=\"za-math-scroll\">\n                    $$ = \\frac{\\frac{1}{5}}{3} \\cdot \\log_{2}2 $$\n                <\/div>\n                \n                Poniewa\u017c \\(\\log_{2}2 = 1\\), zostaje nam samo dzielenie:\n                $$ \\frac{\\frac{1}{5}}{3} = \\frac{1}{5} \\cdot \\frac{1}{3} = \\frac{1}{15} $$\n                \n                <hr class=\"border-success opacity-25 my-4\">\n                <p class=\"fw-bold mb-0 text-success fs-5\">Poprawna odpowied\u017a: A<\/p>\n                \n            <\/div>\n        <\/details>\n\n    <\/div>\n<\/div>\n\n<div class=\"bg-primary bg-opacity-10 border border-primary border-opacity-25 rounded-4 p-4 text-center mt-2\">\n    <p class=\"fw-bold text-primary mb-2\">Gubisz si\u0119 we wzorach na logarytmy?<\/p>\n    <p class=\"small text-muted mb-3\">Zosta\u0142o ma\u0142o czasu do maja, ale z nami na pewno to <strong>Za\u0142apiesz<\/strong>. Przer\u00f3bmy to razem na wirtualnej tablicy.<\/p>\n    <a href=\"https:\/\/zalapto.pl\/#zapisy\" class=\"btn btn-primary btn-sm rounded-pill fw-bold px-4\">Zapisz si\u0119 na korepetycje<\/a>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Wr\u00f3\u0107 do pe\u0142nego arkusza z marcem 2026 Tre\u015b\u0107 zadania Doko\u0144cz zdanie. Wybierz w\u0142a\u015bciw\u0105 odpowied\u017a spo\u015br\u00f3d podanych. Liczba \\(\\log_{8}\\sqrt[5]{2}\\) jest r\u00f3wna: A. \\(\\frac{1}{15}\\) B. \\(-\\frac{1}{15}\\) C. \\(\\frac{3}{5}\\) D. \\(\\frac{5}{3}\\) Rozwi\u0105zanie krok po kroku Zobacz podpowied\u017a Aby obliczy\u0107 ten logarytm bez zgadywania, sprowad\u017a podstaw\u0119 logarytmu (\\(8\\)) i liczb\u0119 logarytmowan\u0105 (\\(\\sqrt[5]{2}\\)) do pot\u0119gi o tej samej podstawie, czyli [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[3,26],"tags":[],"class_list":["post-222","post","type-post","status-publish","format-standard","hentry","category-matura-podstawowa","category-zadania"],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/posts\/222","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/comments?post=222"}],"version-history":[{"count":5,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/posts\/222\/revisions"}],"predecessor-version":[{"id":229,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/posts\/222\/revisions\/229"}],"wp:attachment":[{"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/media?parent=222"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/categories?post=222"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zalapto.pl\/baza\/wp-json\/wp\/v2\/tags?post=222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}