Solving 5^x = 100x – 175 Using Lambert W Function<\/title>\n <style>\n body {\n font-family: 'Georgia', 'Times New Roman', serif;\n line-height: 1.8;\n max-width: 900px;\n margin: 0 auto;\n padding: 40px 20px;\n background: #f9f9f9;\n color: #333;\n }\n \n .container {\n background: white;\n padding: 40px;\n box-shadow: 0 2px 10px rgba(0,0,0,0.1);\n border-radius: 8px;\n }\n \n h1 {\n color: #1a237e;\n text-align: center;\n font-size: 28px;\n margin-bottom: 30px;\n border-bottom: 3px solid #3f51b5;\n padding-bottom: 15px;\n }\n \n h2 {\n color: #283593;\n font-size: 22px;\n margin-top: 35px;\n margin-bottom: 15px;\n border-left: 5px solid #5c6bc0;\n padding-left: 15px;\n }\n \n h3 {\n color: #3949ab;\n font-size: 18px;\n margin-top: 25px;\n margin-bottom: 12px;\n font-style: italic;\n }\n \n p {\n margin: 15px 0;\n text-align: justify;\n }\n \n .equation-block {\n background: #f5f7fa;\n border-left: 4px solid #3f51b5;\n padding: 20px;\n margin: 20px 0;\n border-radius: 4px;\n overflow-x: auto;\n }\n \n .equation {\n text-align: center;\n font-size: 20px;\n margin: 15px 0;\n font-family: 'Cambria Math', 'Times New Roman', serif;\n }\n \n .inline-math {\n font-family: 'Cambria Math', 'Times New Roman', serif;\n font-style: italic;\n }\n \n .fraction {\n display: inline-flex;\n flex-direction: column;\n vertical-align: middle;\n text-align: center;\n margin: 0 4px;\n }\n \n .fraction .numerator {\n border-bottom: 1.5px solid #333;\n padding: 2px 8px;\n font-size: 0.95em;\n }\n \n .fraction .denominator {\n padding: 2px 8px;\n font-size: 0.95em;\n }\n \n sup {\n font-size: 0.75em;\n vertical-align: super;\n }\n \n sub {\n font-size: 0.75em;\n vertical-align: sub;\n }\n \n .highlight-box {\n background: #e8eaf6;\n border: 2px solid #3f51b5;\n padding: 20px;\n margin: 25px 0;\n border-radius: 6px;\n }\n \n .solution-box {\n background: #fff9c4;\n border: 3px solid #fbc02d;\n padding: 25px;\n margin: 30px 0;\n border-radius: 8px;\n box-shadow: 0 2px 8px rgba(251, 192, 45, 0.3);\n }\n \n .final-answer {\n background: #c8e6c9;\n border: 3px solid #4caf50;\n padding: 30px;\n margin: 35px 0;\n border-radius: 8px;\n text-align: center;\n font-size: 24px;\n font-weight: bold;\n color: #1b5e20;\n }\n \n .check-box {\n background: #e8f5e9;\n border-left: 4px solid #4caf50;\n padding: 15px;\n margin: 15px 0;\n border-radius: 4px;\n }\n \n .step-number {\n display: inline-block;\n background: #3f51b5;\n color: white;\n width: 30px;\n height: 30px;\n line-height: 30px;\n text-align: center;\n border-radius: 50%;\n margin-right: 10px;\n font-weight: bold;\n }\n \n .note {\n background: #fff3e0;\n border-left: 4px solid #ff9800;\n padding: 15px;\n margin: 20px 0;\n border-radius: 4px;\n font-style: italic;\n }\n \n .checkmark {\n color: #4caf50;\n font-weight: bold;\n font-size: 1.2em;\n }\n \n table {\n width: 100%;\n margin: 20px 0;\n border-collapse: collapse;\n }\n \n td {\n padding: 10px;\n text-align: center;\n }\n <\/style>\n<\/head>\n<body>\n <div class=\"container\">\n <h1>Solving 5x<\/i><\/sup> = 100x<\/i> \u2212 175 Using the Lambert W Function<\/h1>\n \n \n Objective:<\/strong> Solve the equation systematically using the Lambert W method, without guessing solutions.\n <\/p>\n\n <h2>1<\/span>Rewrite the Equation<\/h2>\n We start with:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 5x<\/i><\/sup> = 100x<\/i> \u2212 175\n <\/div>\n <\/div>\n Using the exponential form 5x<\/i><\/sup> = e<\/i>x<\/i> ln 5<\/sup>:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>x<\/i> ln 5<\/sup> = 100x<\/i> \u2212 175\n <\/div>\n <\/div>\n\n <h2>2<\/span>Eliminate the Constant Term in the Linear Part<\/h2>\n We want the right-hand side to be of the form \u03b1<\/i>(x<\/i> \u2212 \u03b2<\/i>). Factor 100:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 100x<\/i> \u2212 175 = 100(x<\/i> \u2212 \n 175<\/span>\n 100<\/span>\n <\/span>)<\/span> = 100(x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n So the equation becomes:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>x<\/i> ln 5<\/sup> = 100(x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n\n <h2>3<\/span>Shift the Variable<\/h2>\n Let t<\/i> = x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span><\/span>, so x<\/i> = t<\/i> + \n 7<\/span>\n 4<\/span>\n <\/span><\/span>.<\/p>\n \n Substitute into the equation:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>(t<\/i> + \n 7<\/span>\n 4<\/span>\n <\/span>) ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> \u00b7 e<\/i>\n 7<\/span>\n 4<\/span>\n <\/span> ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n 5\n 7<\/span>\n 4<\/span>\n <\/span><\/sup> \u00b7 e<\/i>t<\/i> ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> = \n 100<\/span>\n 57\/4<\/sup><\/span>\n <\/span>t<\/i>\n <\/div>\n <\/div>\n \n Let k<\/i> = \n 100<\/span>\n 57\/4<\/sup><\/span>\n <\/span><\/span>, so:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> = k t<\/i>\n <\/div>\n <\/div>\n\n <h2>4<\/span>Rearrange into Lambert W Form<\/h2>\n Divide both sides by k t<\/i> and rearrange:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 1 = k t e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <div class=\"equation\">\n \n 1<\/span>\n k<\/i><\/span>\n <\/span> = t e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <\/div>\n \n Multiply both sides by \u2212ln 5:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = (\u2212t<\/i> ln 5) e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <\/div>\n \n Let u<\/i> = \u2212t<\/i> ln 5, then:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = u eu<\/sup><\/i>\n <\/div>\n <\/div>\n\n <h2>5<\/span>Apply Lambert W Function<\/h2>\n The Lambert W function is defined as the inverse of f<\/i>(w<\/i>) = w ew<\/sup><\/i>. Therefore:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n u<\/i> = W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <div class=\"equation\">\n t<\/i> = \u2212\n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n \n Since x<\/i> = t<\/i> + 7<\/sup>\u20444<\/sub>:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\" style=\"font-size: 18px;\">\n x<\/i> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n\n <h2>6<\/span>Simplify the Constant Inside W<\/h2>\n Recall k<\/i> = 100<\/sup>\u204457\/4<\/sup><\/sub>. We need to simplify:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = \u2212\n ln 5 \u00b7 57\/4<\/sup><\/span>\n 100<\/span>\n <\/span>\n <\/div>\n <\/div>\n \n Note that 57\/4<\/sup> = 52\u22121\/4<\/sup> = 25 \u00b7 5\u22121\/4<\/sup>. Therefore:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \n ln 5 \u00b7 57\/4<\/sup><\/span>\n 100<\/span>\n <\/span> = \n ln 5 \u00b7 25 \u00b7 5\u22121\/4<\/sup><\/span>\n 100<\/span>\n <\/span> = \n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span>\n <\/div>\n <\/div>\n \n Let A<\/i> = \n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span><\/span>. Then:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\" style=\"font-size: 18px;\">\n x<\/i> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212A<\/i>)\n <\/div>\n <\/div>\n\n <h2>7<\/span>Find W(\u2212A) for Both Real Branches<\/h2>\n Numerically:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n A<\/i> \u2248 \n 1.609438<\/span>\n 4 \u00d7 1.495349<\/span>\n <\/span> \u2248 \n 1.609438<\/span>\n 5.981396<\/span>\n <\/span> \u2248 0.269\n <\/div>\n <\/div>\n \n So \u2212A<\/i> \u2248 \u22120.269, which is greater than \u22121<\/sup>\u2044e<\/i><\/sub> \u2248 \u22120.3679.<\/p>\n \n <div class=\"note\">\n Important:<\/strong> Since \u2212A<\/i> > \u22121<\/sup>\u2044e<\/i><\/sub>, there are two real branches<\/strong> of the Lambert W function: W<\/i>0<\/sub> (principal branch) and W<\/i>\u22121<\/sub> (secondary branch).\n <\/div>\n\n <h2>8<\/span>Solve for W(\u2212A) Exactly<\/h2>\n \n We need W<\/i>(\u2212A<\/i>) such that:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n W<\/i>(\u2212A<\/i>) \u00b7 e<\/i>W<\/i>(\u2212A<\/i>)<\/sup> = \u2212A<\/i> = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span>\n <\/div>\n <\/div>\n\n <h3>First Branch (W0<\/sub>):<\/h3>\n Try W<\/i>0<\/sub>(\u2212A<\/i>) = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n Check:<\/strong><\/p>\n <div class=\"check-box\">\n <div class=\"equation\">\n (\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> \u00b7 exp(\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22121\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span> = \u2212A<\/i> \u2713<\/span>\n <\/div>\n <\/div>\n \n Therefore: W<\/i>0<\/sub>(\u2212A<\/i>) = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n <div class=\"solution-box\">\n <div class=\"equation\" style=\"font-size: 20px; color: #f57c00;\">\n x<\/i>1<\/sub> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 (\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \n 7<\/span>\n 4<\/span>\n <\/span> + \n 1<\/span>\n 4<\/span>\n <\/span> = 2<\/span>\n <\/div>\n <\/div>\n\n <h3>Second Branch (W\u22121<\/sub>):<\/h3>\n Try W<\/i>\u22121<\/sub>(\u2212A<\/i>) = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n Check:<\/strong><\/p>\n <div class=\"check-box\">\n <div class=\"equation\">\n (\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> \u00b7 exp(\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22125\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5 \u00b7 5\u22125\/4<\/sup> = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22121\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span> = \u2212A<\/i> \u2713<\/span>\n <\/div>\n <\/div>\n \n Therefore: W<\/i>\u22121<\/sub>(\u2212A<\/i>) = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n <div class=\"solution-box\">\n <div class=\"equation\" style=\"font-size: 20px; color: #f57c00;\">\n x<\/i>2<\/sub> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 (\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \n 7<\/span>\n 4<\/span>\n <\/span> + \n 5<\/span>\n 4<\/span>\n <\/span> = 3<\/span>\n <\/div>\n <\/div>\n\n <h2>9<\/span>Final Solutions<\/h2>\n <div class=\"final-answer\">\n x<\/i> = 2 and x<\/i> = 3\n <\/div>\n\n <h2>Verification<\/h2>\n <div class=\"highlight-box\">\n <h3>For x<\/i> = 2:<\/h3>\n <table>\n <tr>\n <td>Left side:<\/strong><\/td>\n <td>52<\/sup> = 25<\/td>\n <\/tr>\n <tr>\n <td>Right side:<\/strong><\/td>\n <td>100(2) \u2212 175 = 200 \u2212 175 = 25<\/td>\n <\/tr>\n <tr>\n <td colspan=\"2\" style=\"color: #4caf50; font-weight: bold; font-size: 1.2em;\">25 = 25 \u2713<\/td>\n <\/tr>\n <\/table>\n \n <h3>For x<\/i> = 3:<\/h3>\n <table>\n <tr>\n <td>Left side:<\/strong><\/td>\n <td>53<\/sup> = 125<\/td>\n <\/tr>\n <tr>\n <td>Right side:<\/strong><\/td>\n <td>100(3) \u2212 175 = 300 \u2212 175 = 125<\/td>\n <\/tr>\n <tr>\n <td colspan=\"2\" style=\"color: #4caf50; font-weight: bold; font-size: 1.2em;\">125 = 125 \u2713<\/td>\n <\/tr>\n <\/table>\n <\/div>\n\n \n This solution demonstrates the power of the Lambert W function in solving transcendental equations that combine exponential and linear terms.\n <\/p>\n <\/div>\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"how do you know this is W sub zero (W \u22120) or W sub -1 (W \u22121), in other words how do you distinguish between the two real branches in the above context \u200b \u200b Excellent question. The distinction between W 0 W 0 \u200b and W \u2212 1 W \u22121 \u200b is determined by […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_themeisle_gutenberg_block_has_review":false,"footnotes":""},"class_list":["post-39","page","type-page","status-publish","hentry"],"yoast_head":"\n<title>Lambert W Function FAQ - Math Lab<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Lambert W Function FAQ - Math Lab\" \/>\n<meta property=\"og:description\" content=\"how do you know this is W sub zero (W \u22120) or W sub -1 (W \u22121), in other words how do you distinguish between the two real branches in the above context \u200b \u200b Excellent question. 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The distinction between W 0 W 0 \u200b and W \u2212 1 W \u22121 \u200b is determined by […]","og_url":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/","og_site_name":"Math Lab","article_modified_time":"2025-11-08T19:45:57+00:00","twitter_card":"summary_large_image","twitter_misc":{"Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/","url":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/","name":"Lambert W Function FAQ - Math Lab","isPartOf":{"@id":"https:\/\/potentsky.com\/math\/#website"},"datePublished":"2025-11-08T10:45:13+00:00","dateModified":"2025-11-08T19:45:57+00:00","breadcrumb":{"@id":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/potentsky.com\/math\/lambert-w-function-faq\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/potentsky.com\/math\/"},{"@type":"ListItem","position":2,"name":"Lambert W Function FAQ"}]},{"@type":"WebSite","@id":"https:\/\/potentsky.com\/math\/#website","url":"https:\/\/potentsky.com\/math\/","name":"Math Lab","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/potentsky.com\/math\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"}]}},"_links":{"self":[{"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/pages\/39","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/comments?post=39"}],"version-history":[{"count":3,"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/pages\/39\/revisions"}],"predecessor-version":[{"id":48,"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/pages\/39\/revisions\/48"}],"wp:attachment":[{"href":"https:\/\/potentsky.com\/math\/wp-json\/wp\/v2\/media?parent=39"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":39,"date":"2025-11-08T10:45:13","date_gmt":"2025-11-08T10:45:13","guid":{"rendered":"https:\/\/potentsky.com\/math\/?page_id=39"},"modified":"2025-11-08T19:45:57","modified_gmt":"2025-11-08T19:45:57","slug":"lambert-w-function-faq","status":"publish","type":"page","link":"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/","title":{"rendered":"Lambert W Function FAQ"},"content":{"rendered":"\nhow do you know this is W sub zero (W \u22120) or W sub -1 (W \u22121), in other words how do you distinguish between the two real branches in the above context\n\u200b\n \n\n\u200b\n\n\n\nExcellent question. The distinction between \nW\n0\nW \n0\n\u200b\n and \nW\n\u2212\n1\nW \n\u22121\n\u200b\n is determined by the range of the function \nW\n(\nz\n)\nW(z) for real \nz\nz.\n\n\n\n\n\n\n \n \n Solving 5^x = 100x – 175 Using Lambert W Function<\/title>\n <style>\n body {\n font-family: 'Georgia', 'Times New Roman', serif;\n line-height: 1.8;\n max-width: 900px;\n margin: 0 auto;\n padding: 40px 20px;\n background: #f9f9f9;\n color: #333;\n }\n \n .container {\n background: white;\n padding: 40px;\n box-shadow: 0 2px 10px rgba(0,0,0,0.1);\n border-radius: 8px;\n }\n \n h1 {\n color: #1a237e;\n text-align: center;\n font-size: 28px;\n margin-bottom: 30px;\n border-bottom: 3px solid #3f51b5;\n padding-bottom: 15px;\n }\n \n h2 {\n color: #283593;\n font-size: 22px;\n margin-top: 35px;\n margin-bottom: 15px;\n border-left: 5px solid #5c6bc0;\n padding-left: 15px;\n }\n \n h3 {\n color: #3949ab;\n font-size: 18px;\n margin-top: 25px;\n margin-bottom: 12px;\n font-style: italic;\n }\n \n p {\n margin: 15px 0;\n text-align: justify;\n }\n \n .equation-block {\n background: #f5f7fa;\n border-left: 4px solid #3f51b5;\n padding: 20px;\n margin: 20px 0;\n border-radius: 4px;\n overflow-x: auto;\n }\n \n .equation {\n text-align: center;\n font-size: 20px;\n margin: 15px 0;\n font-family: 'Cambria Math', 'Times New Roman', serif;\n }\n \n .inline-math {\n font-family: 'Cambria Math', 'Times New Roman', serif;\n font-style: italic;\n }\n \n .fraction {\n display: inline-flex;\n flex-direction: column;\n vertical-align: middle;\n text-align: center;\n margin: 0 4px;\n }\n \n .fraction .numerator {\n border-bottom: 1.5px solid #333;\n padding: 2px 8px;\n font-size: 0.95em;\n }\n \n .fraction .denominator {\n padding: 2px 8px;\n font-size: 0.95em;\n }\n \n sup {\n font-size: 0.75em;\n vertical-align: super;\n }\n \n sub {\n font-size: 0.75em;\n vertical-align: sub;\n }\n \n .highlight-box {\n background: #e8eaf6;\n border: 2px solid #3f51b5;\n padding: 20px;\n margin: 25px 0;\n border-radius: 6px;\n }\n \n .solution-box {\n background: #fff9c4;\n border: 3px solid #fbc02d;\n padding: 25px;\n margin: 30px 0;\n border-radius: 8px;\n box-shadow: 0 2px 8px rgba(251, 192, 45, 0.3);\n }\n \n .final-answer {\n background: #c8e6c9;\n border: 3px solid #4caf50;\n padding: 30px;\n margin: 35px 0;\n border-radius: 8px;\n text-align: center;\n font-size: 24px;\n font-weight: bold;\n color: #1b5e20;\n }\n \n .check-box {\n background: #e8f5e9;\n border-left: 4px solid #4caf50;\n padding: 15px;\n margin: 15px 0;\n border-radius: 4px;\n }\n \n .step-number {\n display: inline-block;\n background: #3f51b5;\n color: white;\n width: 30px;\n height: 30px;\n line-height: 30px;\n text-align: center;\n border-radius: 50%;\n margin-right: 10px;\n font-weight: bold;\n }\n \n .note {\n background: #fff3e0;\n border-left: 4px solid #ff9800;\n padding: 15px;\n margin: 20px 0;\n border-radius: 4px;\n font-style: italic;\n }\n \n .checkmark {\n color: #4caf50;\n font-weight: bold;\n font-size: 1.2em;\n }\n \n table {\n width: 100%;\n margin: 20px 0;\n border-collapse: collapse;\n }\n \n td {\n padding: 10px;\n text-align: center;\n }\n <\/style>\n<\/head>\n<body>\n <div class=\"container\">\n <h1>Solving 5x<\/i><\/sup> = 100x<\/i> \u2212 175 Using the Lambert W Function<\/h1>\n \n \n Objective:<\/strong> Solve the equation systematically using the Lambert W method, without guessing solutions.\n <\/p>\n\n <h2>1<\/span>Rewrite the Equation<\/h2>\n We start with:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 5x<\/i><\/sup> = 100x<\/i> \u2212 175\n <\/div>\n <\/div>\n Using the exponential form 5x<\/i><\/sup> = e<\/i>x<\/i> ln 5<\/sup>:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>x<\/i> ln 5<\/sup> = 100x<\/i> \u2212 175\n <\/div>\n <\/div>\n\n <h2>2<\/span>Eliminate the Constant Term in the Linear Part<\/h2>\n We want the right-hand side to be of the form \u03b1<\/i>(x<\/i> \u2212 \u03b2<\/i>). Factor 100:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 100x<\/i> \u2212 175 = 100(x<\/i> \u2212 \n 175<\/span>\n 100<\/span>\n <\/span>)<\/span> = 100(x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n So the equation becomes:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>x<\/i> ln 5<\/sup> = 100(x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n\n <h2>3<\/span>Shift the Variable<\/h2>\n Let t<\/i> = x<\/i> \u2212 \n 7<\/span>\n 4<\/span>\n <\/span><\/span>, so x<\/i> = t<\/i> + \n 7<\/span>\n 4<\/span>\n <\/span><\/span>.<\/p>\n \n Substitute into the equation:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n e<\/i>(t<\/i> + \n 7<\/span>\n 4<\/span>\n <\/span>) ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> \u00b7 e<\/i>\n 7<\/span>\n 4<\/span>\n <\/span> ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n 5\n 7<\/span>\n 4<\/span>\n <\/span><\/sup> \u00b7 e<\/i>t<\/i> ln 5<\/sup> = 100t<\/i>\n <\/div>\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> = \n 100<\/span>\n 57\/4<\/sup><\/span>\n <\/span>t<\/i>\n <\/div>\n <\/div>\n \n Let k<\/i> = \n 100<\/span>\n 57\/4<\/sup><\/span>\n <\/span><\/span>, so:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\">\n e<\/i>t<\/i> ln 5<\/sup> = k t<\/i>\n <\/div>\n <\/div>\n\n <h2>4<\/span>Rearrange into Lambert W Form<\/h2>\n Divide both sides by k t<\/i> and rearrange:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n 1 = k t e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <div class=\"equation\">\n \n 1<\/span>\n k<\/i><\/span>\n <\/span> = t e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <\/div>\n \n Multiply both sides by \u2212ln 5:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = (\u2212t<\/i> ln 5) e<\/i>\u2212t<\/i> ln 5<\/sup>\n <\/div>\n <\/div>\n \n Let u<\/i> = \u2212t<\/i> ln 5, then:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = u eu<\/sup><\/i>\n <\/div>\n <\/div>\n\n <h2>5<\/span>Apply Lambert W Function<\/h2>\n The Lambert W function is defined as the inverse of f<\/i>(w<\/i>) = w ew<\/sup><\/i>. Therefore:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n u<\/i> = W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <div class=\"equation\">\n t<\/i> = \u2212\n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n \n Since x<\/i> = t<\/i> + 7<\/sup>\u20444<\/sub>:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\" style=\"font-size: 18px;\">\n x<\/i> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span>)<\/span>\n <\/div>\n <\/div>\n\n <h2>6<\/span>Simplify the Constant Inside W<\/h2>\n Recall k<\/i> = 100<\/sup>\u204457\/4<\/sup><\/sub>. We need to simplify:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \u2212\n ln 5<\/span>\n k<\/i><\/span>\n <\/span> = \u2212\n ln 5 \u00b7 57\/4<\/sup><\/span>\n 100<\/span>\n <\/span>\n <\/div>\n <\/div>\n \n Note that 57\/4<\/sup> = 52\u22121\/4<\/sup> = 25 \u00b7 5\u22121\/4<\/sup>. Therefore:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n \n ln 5 \u00b7 57\/4<\/sup><\/span>\n 100<\/span>\n <\/span> = \n ln 5 \u00b7 25 \u00b7 5\u22121\/4<\/sup><\/span>\n 100<\/span>\n <\/span> = \n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span>\n <\/div>\n <\/div>\n \n Let A<\/i> = \n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span><\/span>. Then:<\/p>\n <div class=\"highlight-box\">\n <div class=\"equation\" style=\"font-size: 18px;\">\n x<\/i> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 W<\/i>(\u2212A<\/i>)\n <\/div>\n <\/div>\n\n <h2>7<\/span>Find W(\u2212A) for Both Real Branches<\/h2>\n Numerically:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n A<\/i> \u2248 \n 1.609438<\/span>\n 4 \u00d7 1.495349<\/span>\n <\/span> \u2248 \n 1.609438<\/span>\n 5.981396<\/span>\n <\/span> \u2248 0.269\n <\/div>\n <\/div>\n \n So \u2212A<\/i> \u2248 \u22120.269, which is greater than \u22121<\/sup>\u2044e<\/i><\/sub> \u2248 \u22120.3679.<\/p>\n \n <div class=\"note\">\n Important:<\/strong> Since \u2212A<\/i> > \u22121<\/sup>\u2044e<\/i><\/sub>, there are two real branches<\/strong> of the Lambert W function: W<\/i>0<\/sub> (principal branch) and W<\/i>\u22121<\/sub> (secondary branch).\n <\/div>\n\n <h2>8<\/span>Solve for W(\u2212A) Exactly<\/h2>\n \n We need W<\/i>(\u2212A<\/i>) such that:<\/p>\n <div class=\"equation-block\">\n <div class=\"equation\">\n W<\/i>(\u2212A<\/i>) \u00b7 e<\/i>W<\/i>(\u2212A<\/i>)<\/sup> = \u2212A<\/i> = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span>\n <\/div>\n <\/div>\n\n <h3>First Branch (W0<\/sub>):<\/h3>\n Try W<\/i>0<\/sub>(\u2212A<\/i>) = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n Check:<\/strong><\/p>\n <div class=\"check-box\">\n <div class=\"equation\">\n (\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> \u00b7 exp(\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22121\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span> = \u2212A<\/i> \u2713<\/span>\n <\/div>\n <\/div>\n \n Therefore: W<\/i>0<\/sub>(\u2212A<\/i>) = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n <div class=\"solution-box\">\n <div class=\"equation\" style=\"font-size: 20px; color: #f57c00;\">\n x<\/i>1<\/sub> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 (\u2212\n ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \n 7<\/span>\n 4<\/span>\n <\/span> + \n 1<\/span>\n 4<\/span>\n <\/span> = 2<\/span>\n <\/div>\n <\/div>\n\n <h3>Second Branch (W\u22121<\/sub>):<\/h3>\n Try W<\/i>\u22121<\/sub>(\u2212A<\/i>) = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n Check:<\/strong><\/p>\n <div class=\"check-box\">\n <div class=\"equation\">\n (\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> \u00b7 exp(\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22125\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5 \u00b7 5\u22125\/4<\/sup> = \u2212\n ln 5<\/span>\n 4<\/span>\n <\/span> \u00b7 5\u22121\/4<\/sup>\n <\/div>\n <div class=\"equation\">\n = \u2212\n ln 5<\/span>\n 4 \u00b7 51\/4<\/sup><\/span>\n <\/span> = \u2212A<\/i> \u2713<\/span>\n <\/div>\n <\/div>\n \n Therefore: W<\/i>\u22121<\/sub>(\u2212A<\/i>) = \u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span><\/p>\n \n <div class=\"solution-box\">\n <div class=\"equation\" style=\"font-size: 20px; color: #f57c00;\">\n x<\/i>2<\/sub> = \n 7<\/span>\n 4<\/span>\n <\/span> \u2212 \n 1<\/span>\n ln 5<\/span>\n <\/span> \u00b7 (\u2212\n 5 ln 5<\/span>\n 4<\/span>\n <\/span>)<\/span> = \n 7<\/span>\n 4<\/span>\n <\/span> + \n 5<\/span>\n 4<\/span>\n <\/span> = 3<\/span>\n <\/div>\n <\/div>\n\n <h2>9<\/span>Final Solutions<\/h2>\n <div class=\"final-answer\">\n x<\/i> = 2 and x<\/i> = 3\n <\/div>\n\n <h2>Verification<\/h2>\n <div class=\"highlight-box\">\n <h3>For x<\/i> = 2:<\/h3>\n <table>\n <tr>\n <td>Left side:<\/strong><\/td>\n <td>52<\/sup> = 25<\/td>\n <\/tr>\n <tr>\n <td>Right side:<\/strong><\/td>\n <td>100(2) \u2212 175 = 200 \u2212 175 = 25<\/td>\n <\/tr>\n <tr>\n <td colspan=\"2\" style=\"color: #4caf50; font-weight: bold; font-size: 1.2em;\">25 = 25 \u2713<\/td>\n <\/tr>\n <\/table>\n \n <h3>For x<\/i> = 3:<\/h3>\n <table>\n <tr>\n <td>Left side:<\/strong><\/td>\n <td>53<\/sup> = 125<\/td>\n <\/tr>\n <tr>\n <td>Right side:<\/strong><\/td>\n <td>100(3) \u2212 175 = 300 \u2212 175 = 125<\/td>\n <\/tr>\n <tr>\n <td colspan=\"2\" style=\"color: #4caf50; font-weight: bold; font-size: 1.2em;\">125 = 125 \u2713<\/td>\n <\/tr>\n <\/table>\n <\/div>\n\n \n This solution demonstrates the power of the Lambert W function in solving transcendental equations that combine exponential and linear terms.\n <\/p>\n <\/div>\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"how do you know this is W sub zero (W \u22120) or W sub -1 (W \u22121), in other words how do you distinguish between the two real branches in the above context \u200b \u200b Excellent question. The distinction between W 0 W 0 \u200b and W \u2212 1 W \u22121 \u200b is determined by […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_themeisle_gutenberg_block_has_review":false,"footnotes":""},"class_list":["post-39","page","type-page","status-publish","hentry"],"yoast_head":"\n<title>Lambert W Function FAQ - Math Lab<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/potentsky.com\/math\/lambert-w-function-faq\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Lambert W Function FAQ - Math Lab\" \/>\n<meta property=\"og:description\" content=\"how do you know this is W sub zero (W \u22120) or W sub -1 (W \u22121), in other words how do you distinguish between the two real branches in the above context \u200b \u200b Excellent question. 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