{"id":221608,"date":"2025-05-29T20:29:18","date_gmt":"2025-05-29T20:29:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221608"},"modified":"2025-05-29T20:29:20","modified_gmt":"2025-05-29T20:29:20","slug":"logg1-32-14-log-1-27-15-log-in-problems-17-28-solve-the-given-equation-for-x-wit-17","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/logg1-32-14-log-1-27-15-log-in-problems-17-28-solve-the-given-equation-for-x-wit-17\/","title":{"rendered":"logg(1\/32). 14. log (1\/27). 15. log In problems 17-28, solve the given equation for x wit 17"},"content":{"rendered":"\n

logg(1\/32). 14. log (1\/27). 15. log In problems 17-28, solve the given equation for x wit 17. log, (27)–3. 20. log4 (x \/3)–5\/2. 23) log3 (2x-3)-2 26. 10g 12(2x-1)-2 18, log, (2) 3. 21. log32 (3x) -3\/5. 24, log2(1-5x) 4. 27 log2(3×2+2x 1 In problems 13-32, graph the given function by us the graph of the basic functions F(x) l09x, F(x) 29. f(x)- logd-3. 2 2<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Problem Solutions:<\/strong><\/h3>\n\n\n\n

13.<\/strong> log\u2061(1\/32)\\log(1\/32)
log\u2061(1\/32)=log\u2061(32\u22121)=\u2212log\u2061(32)\\log(1\/32) = \\log(32^{-1}) = -\\log(32)
Since 32=2532 = 2^5,
\u2212log\u2061(32)=\u2212log\u2061(25)=\u22125log\u2061(2)\u2248\u22125(0.3010)=\u22121.505-\\log(32) = -\\log(2^5) = -5\\log(2) \\approx -5(0.3010) = -1.505<\/p>\n\n\n\n

\n\n\n\n

14.<\/strong> log\u2061(1\/27)\\log(1\/27)
log\u2061(1\/27)=\u2212log\u2061(27)=\u2212log\u2061(33)=\u22123log\u2061(3)\u2248\u22123(0.4771)=\u22121.4313\\log(1\/27) = -\\log(27) = -\\log(3^3) = -3\\log(3) \\approx -3(0.4771) = -1.4313<\/p>\n\n\n\n

\n\n\n\n

15.<\/strong> The question appears incomplete.<\/p>\n\n\n\n

\n\n\n\n

17.<\/strong> log\u2061b(27)=\u22123\\log_b(27) = -3
This means:
b\u22123=27\u21d21b3=27\u21d2b3=127\u21d2b=13b^{-3} = 27 \\Rightarrow \\frac{1}{b^3} = 27 \\Rightarrow b^3 = \\frac{1}{27} \\Rightarrow b = \\frac{1}{3}<\/p>\n\n\n\n

\n\n\n\n

18.<\/strong> log\u2061b(2)=3\\log_b(2) = 3
So,
b3=2\u21d2b=23\u22481.26b^3 = 2 \\Rightarrow b = \\sqrt[3]{2} \\approx 1.26<\/p>\n\n\n\n

\n\n\n\n

20.<\/strong> log\u20614(x\/3)=\u22125\/2\\log_4(x\/3) = -5\/2
x\/3=4\u22125\/2=145\/2=132\u21d2x=3\/32x\/3 = 4^{-5\/2} = \\frac{1}{4^{5\/2}} = \\frac{1}{32} \\Rightarrow x = 3\/32<\/p>\n\n\n\n

\n\n\n\n

21.<\/strong> log\u206132(3x)=\u22123\/5\\log_{32}(3x) = -3\/5
Rewrite:
3x=32\u22123\/5=(25)\u22123\/5=2\u22123=1\/8\u21d2x=1\/243x = 32^{-3\/5} = (2^5)^{-3\/5} = 2^{-3} = 1\/8 \\Rightarrow x = 1\/24<\/p>\n\n\n\n

\n\n\n\n

23.<\/strong> log\u20613(2x\u22123)=2\u21d22x\u22123=32=9\u21d2x=6\\log_3(2x – 3) = 2 \\Rightarrow 2x – 3 = 3^2 = 9 \\Rightarrow x = 6<\/p>\n\n\n\n

\n\n\n\n

24.<\/strong> log\u20612(1\u22125x)=4\u21d21\u22125x=24=16\u21d2x=\u22123\\log_2(1 – 5x) = 4 \\Rightarrow 1 – 5x = 2^4 = 16 \\Rightarrow x = -3<\/p>\n\n\n\n

\n\n\n\n

26.<\/strong> log\u206112(2x\u22121)=\u22122\u21d22x\u22121=12\u22122=1\/144\u21d2x=145\/288\\log_{12}(2x – 1) = -2 \\Rightarrow 2x – 1 = 12^{-2} = 1\/144 \\Rightarrow x = 145\/288<\/p>\n\n\n\n

\n\n\n\n

27.<\/strong> log\u20612(3×2+2x+1)\\log_2(3x^2 + 2x + 1)
This is not an equation, so no solution for xx; it may be for graphing.<\/p>\n\n\n\n

\n\n\n\n

29.<\/strong> f(x)=log\u2061(x\u22123)f(x) = \\log(x – 3)
This is a transformation of the parent function f(x)=log\u2061xf(x) = \\log x. It’s shifted 3 units to the right<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Explanation:<\/strong><\/h3>\n\n\n\n

Logarithms are the inverse of exponentiation. The expression log\u2061b(a)=c\\log_b(a) = c means bc=ab^c = a. They are used to solve exponential equations and simplify multiplicative processes. Common bases include base 10 (common logarithms) and base e\u22482.718e \\approx 2.718 (natural logarithms).<\/p>\n\n\n\n

For logarithmic evaluations like Problems 13 and 14<\/strong>, we apply properties such as:<\/p>\n\n\n\n

    \n
  • log\u2061(1\/a)=\u2212log\u2061(a)\\log(1\/a) = -\\log(a)<\/li>\n\n\n\n
  • log\u2061(ab)=blog\u2061(a)\\log(a^b) = b\\log(a)<\/li>\n<\/ul>\n\n\n\n

    This helps simplify logs with fractional arguments. For example, log\u2061(1\/32)=\u2212log\u2061(32)=\u22125log\u2061(2)\\log(1\/32) = -\\log(32) = -5\\log(2).<\/p>\n\n\n\n

    In solving equations like 17 to 26<\/strong>, we use the definition of logarithms to rewrite the equation in exponential form. For example, log\u2061b(a)=c\\log_b(a) = c means bc=ab^c = a. Solving for xx often involves algebraic manipulation after converting to exponential form.<\/p>\n\n\n\n

    Domain is important:<\/strong> The argument of a logarithmic function must be positive. For instance, in log\u20612(1\u22125x)=4\\log_2(1 – 5x) = 4, we must check that 1\u22125x>0\u21d2x<0.21 – 5x > 0 \\Rightarrow x < 0.2 \u2014 this confirms the solution x=\u22123x = -3 is valid.<\/p>\n\n\n\n

    For graphing functions like 29<\/strong>, we use transformations. The basic graph y=log\u2061xy = \\log x is shifted right by 3 units to become y=log\u2061(x\u22123)y = \\log(x – 3). This shift changes the vertical asymptote from x=0x = 0 to x=3x = 3.<\/p>\n\n\n\n

    Understanding these properties is essential in algebra, calculus, and applications like data modeling and sound intensity in physics.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    logg(1\/32). 14. log (1\/27). 15. log In problems 17-28, solve the given equation for x wit 17. log, (27)–3. 20. log4 (x \/3)–5\/2. 23) log3 (2x-3)-2 26. 10g 12(2x-1)-2 18, log, (2) 3. 21. log32 (3x) -3\/5. 24, log2(1-5x) 4. 27 log2(3×2+2x 1 In problems 13-32, graph the given function by us the graph of […]<\/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 center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-221608","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221608","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=221608"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221608\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221608"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221608"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221608"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}