Logarithms Practice Test 1

© examsiri.com

Question : 27 of 40

Marks: +1, -0

If

\log_12 27 = a

, then the value of

\log_6 16

is:

${3+a}/{4(3-a)}$
${3-a}/{4(3+a)}$
${4(3+a)}/{(3-a)}$
${4(3-a)}/{(3+a)}$

Solution:

\log_12 27 = a

⇒{\log27}/{\log12}=a

⇒{\log3^3}/{\log(2^2×3)}=a

⇒{3\log3 }/{2\log 2 +\log3 }=a

⇒{2\log 2 +\log3 }/{3\log3 }=1/a

⇒{2\log 2}/{2\log3 } +{\log3 }/{3\log3 }=1/a

⇒{2\log 2}/{3\log3 } =1/a-1/3

⇒{2\log 2}/{3\log3 } = {3-a}/{3a}

⇒\log3={2a}/(3-a)\log 2

Now,

\log_6 16={\log 2^4}/{\log(2×3)}

={4\log2}/{\log2+\log3}

={4\log2}/{\log2+{2a}/(3-a)\log 2}

={4(3-a)}/{3+a}

© examsiri.com

Go to Question:

Prev Question

Next Question