{"id":234549,"date":"2025-06-14T08:21:59","date_gmt":"2025-06-14T08:21:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234549"},"modified":"2025-06-14T08:22:01","modified_gmt":"2025-06-14T08:22:01","slug":"log-base-8-0-25-compute-the-log","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/log-base-8-0-25-compute-the-log\/","title":{"rendered":"log base 8 (0.25) compute the log"},"content":{"rendered":"\n<p>log base 8 (0.25) compute the log<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To compute:log\u20618(0.25)\\log_8(0.25)log8\u200b(0.25)<\/p>\n\n\n\n<p><strong>Step 1: Change of base formula<\/strong><br>Use the logarithmic identity:log\u2061b(a)=log\u2061c(a)log\u2061c(b)\\log_b(a) = \\frac{\\log_c(a)}{\\log_c(b)}logb\u200b(a)=logc\u200b(b)logc\u200b(a)\u200b<\/p>\n\n\n\n<p>Let base c=10c = 10c=10:log\u20618(0.25)=log\u206110(0.25)log\u206110(8)\\log_8(0.25) = \\frac{\\log_{10}(0.25)}{\\log_{10}(8)}log8\u200b(0.25)=log10\u200b(8)log10\u200b(0.25)\u200b<\/p>\n\n\n\n<p><strong>Step 2: Calculate the values<\/strong><br>Use a calculator or log tables:log\u206110(0.25)\u2248\u22120.6021log\u206110(8)=log\u206110(23)=3log\u206110(2)\u22483\u00d70.3010=0.9030\\log_{10}(0.25) \\approx -0.6021 \\\\ \\log_{10}(8) = \\log_{10}(2^3) = 3\\log_{10}(2) \\approx 3 \\times 0.3010 = 0.9030log10\u200b(0.25)\u2248\u22120.6021log10\u200b(8)=log10\u200b(23)=3log10\u200b(2)\u22483\u00d70.3010=0.9030<\/p>\n\n\n\n<p>Now divide:log\u20618(0.25)\u2248\u22120.60210.9030\u2248\u22120.6667\\log_8(0.25) \\approx \\frac{-0.6021}{0.9030} \\approx -0.6667log8\u200b(0.25)\u22480.9030\u22120.6021\u200b\u2248\u22120.6667<\/p>\n\n\n\n<p><strong>Final Answer:<\/strong>log\u20618(0.25)\u2248\u221223\\log_8(0.25) \\approx -\\frac{2}{3}log8\u200b(0.25)\u2248\u221232\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Explanation<\/strong><\/p>\n\n\n\n<p>The logarithmic expression log\u20618(0.25)\\log_8(0.25)log8\u200b(0.25) asks for the exponent to which 8 must be raised to yield 0.25. Since 8 is greater than 1 and 0.25 is less than 1, the result must be negative.<\/p>\n\n\n\n<p>To simplify computation, the change-of-base formula proves useful. This identity allows conversion of a logarithm with any base into a ratio of logarithms with a common base such as 10 or eee. Applying this identity:log\u20618(0.25)=log\u206110(0.25)log\u206110(8)\\log_8(0.25) = \\frac{\\log_{10}(0.25)}{\\log_{10}(8)}log8\u200b(0.25)=log10\u200b(8)log10\u200b(0.25)\u200b<\/p>\n\n\n\n<p>Using decimal approximations:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>log\u206110(0.25)\u2248\u22120.6021\\log_{10}(0.25) \\approx -0.6021log10\u200b(0.25)\u2248\u22120.6021, since 0.25 equals 14\\frac{1}{4}41\u200b, and logarithms of values between 0 and 1 are negative.<\/li>\n\n\n\n<li>log\u206110(8)=log\u206110(23)=3log\u206110(2)\u22483\u00d70.3010=0.9030\\log_{10}(8) = \\log_{10}(2^3) = 3\\log_{10}(2) \\approx 3 \\times 0.3010 = 0.9030log10\u200b(8)=log10\u200b(23)=3log10\u200b(2)\u22483\u00d70.3010=0.9030<\/li>\n<\/ul>\n\n\n\n<p>Thus,log\u20618(0.25)\u2248\u22120.60210.9030\u2248\u22120.6667\\log_8(0.25) \\approx \\frac{-0.6021}{0.9030} \\approx -0.6667log8\u200b(0.25)\u22480.9030\u22120.6021\u200b\u2248\u22120.6667<\/p>\n\n\n\n<p>The decimal \u22120.6667-0.6667\u22120.6667 equals the fraction \u221223-\\frac{2}{3}\u221232\u200b. This means that:8\u22122\/3=0.258^{-2\/3} = 0.258\u22122\/3=0.25<\/p>\n\n\n\n<p>To verify, rewrite 8 as 232^323:(23)\u22122\/3=2\u22122=14=0.25(2^3)^{-2\/3} = 2^{-2} = \\frac{1}{4} = 0.25(23)\u22122\/3=2\u22122=41\u200b=0.25<\/p>\n\n\n\n<p>This confirms the result is exact. The logarithm evaluates to \u221223-\\frac{2}{3}\u221232\u200b.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-435.jpeg\" alt=\"\" class=\"wp-image-234550\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>log base 8 (0.25) compute the log The Correct Answer and Explanation is: To compute:log\u20618(0.25)\\log_8(0.25)log8\u200b(0.25) Step 1: Change of base formulaUse the logarithmic identity:log\u2061b(a)=log\u2061c(a)log\u2061c(b)\\log_b(a) = \\frac{\\log_c(a)}{\\log_c(b)}logb\u200b(a)=logc\u200b(b)logc\u200b(a)\u200b Let base c=10c = 10c=10:log\u20618(0.25)=log\u206110(0.25)log\u206110(8)\\log_8(0.25) = \\frac{\\log_{10}(0.25)}{\\log_{10}(8)}log8\u200b(0.25)=log10\u200b(8)log10\u200b(0.25)\u200b Step 2: Calculate the valuesUse a calculator or log tables:log\u206110(0.25)\u2248\u22120.6021log\u206110(8)=log\u206110(23)=3log\u206110(2)\u22483\u00d70.3010=0.9030\\log_{10}(0.25) \\approx -0.6021 \\\\ \\log_{10}(8) = \\log_{10}(2^3) = 3\\log_{10}(2) \\approx 3 \\times 0.3010 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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