Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb.<\/p>\n\n\n\n
log6(2)=
log6(8\/6)=<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n We are given the logarithmic values:<\/p>\n\n\n\n [ [ Using the change of base formula<\/strong>:<\/p>\n\n\n\n [ we set ( k = 2 ) to express ( \\log_6(2) ) in terms of base 2 logarithms:<\/p>\n\n\n\n [ Since ( \\log_2(2) = 1 ), we substitute:<\/p>\n\n\n\n [ Since ( \\log_2(6) = b ), we get:<\/p>\n\n\n\n [ Using the quotient rule<\/strong> for logarithms:<\/p>\n\n\n\n [ we apply this to ( \\log_6(8\/6) ):<\/p>\n\n\n\n [ Using the change of base formula<\/strong>:<\/p>\n\n\n\n [ [ Thus:<\/p>\n\n\n\n [ [ To solve these logarithmic expressions in terms of ( a ) and ( b ), we use logarithmic properties. The change of base formula<\/strong> allows us to convert logarithms of any base into logarithms of base 2, which we use because we are given ( \\log_2(8) ) and ( \\log_2(6) ).<\/p>\n\n\n\n For ( \\log_6(2) ), applying the change of base formula gives:<\/p>\n\n\n\n [ Since we know that ( \\log_2(2) = 1 ) and ( \\log_2(6) = b ), the result is simply ( \\frac{1}{b} ).<\/p>\n\n\n\n Next, for ( \\log_6(8\/6) ), we use the quotient rule<\/strong> for logarithms:<\/p>\n\n\n\n [ Using the change of base formula, we convert ( \\log_6(8) ) and ( \\log_6(6) ) into base 2:<\/p>\n\n\n\n [ [ Thus, we obtain:<\/p>\n\n\n\n [ These results show how logarithmic identities and base change techniques simplify complex logarithmic expressions.<\/p>\n","protected":false},"excerpt":{"rendered":" Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb. log6(2)=log6(8\/6)= The correct answer and explanation is : We are given the logarithmic values: [\\log_2(8) = a] [\\log_2(6) = b] Step 1: Expressing ( \\log_6(2) ) in Terms 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 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\\log_2(8) = a
]<\/p>\n\n\n\n
\\log_2(6) = b
]<\/p>\n\n\n\nStep 1: Expressing ( \\log_6(2) ) in Terms of ( a ) and ( b )<\/strong><\/h3>\n\n\n\n
\\log_b(x) = \\frac{\\log_k(x)}{\\log_k(b)}
]<\/p>\n\n\n\n
\\log_6(2) = \\frac{\\log_2(2)}{\\log_2(6)}
]<\/p>\n\n\n\n
\\log_6(2) = \\frac{1}{\\log_2(6)}
]<\/p>\n\n\n\n
\\log_6(2) = \\frac{1}{b}
]<\/p>\n\n\n\n
\n\n\n\nStep 2: Expressing ( \\log_6(8\/6) ) in Terms of ( a ) and ( b )<\/strong><\/h3>\n\n\n\n
\\log_b(x\/y) = \\log_b(x) – \\log_b(y)
]<\/p>\n\n\n\n
\\log_6(8\/6) = \\log_6(8) – \\log_6(6)
]<\/p>\n\n\n\n
\\log_6(8) = \\frac{\\log_2(8)}{\\log_2(6)} = \\frac{a}{b}
]<\/p>\n\n\n\n
\\log_6(6) = \\frac{\\log_2(6)}{\\log_2(6)} = 1
]<\/p>\n\n\n\n
\\log_6(8\/6) = \\frac{a}{b} – 1
]<\/p>\n\n\n\n
\n\n\n\nFinal Answers:<\/strong><\/h3>\n\n\n\n
\\log_6(2) = \\frac{1}{b}
]
[
\\log_6(8\/6) = \\frac{a}{b} – 1
]<\/p>\n\n\n\nExplanation (300 words)<\/strong><\/h3>\n\n\n\n
\\log_6(2) = \\frac{\\log_2(2)}{\\log_2(6)}
]<\/p>\n\n\n\n
\\log_6(8\/6) = \\log_6(8) – \\log_6(6)
]<\/p>\n\n\n\n
\\log_6(8) = \\frac{\\log_2(8)}{\\log_2(6)} = \\frac{a}{b}
]<\/p>\n\n\n\n
\\log_6(6) = \\frac{\\log_2(6)}{\\log_2(6)} = 1
]<\/p>\n\n\n\n
\\log_6(8\/6) = \\frac{a}{b} – 1
]<\/p>\n\n\n\n