Question

if 1/2 log(base 3)m + 3 log(base a) n = 1,express m in terms of n. please solve it's urgent





answer : m=9 (base n) ^-6​

Answer 1

Answer:

$\large\boxed{\sf{m = \dfrac{9}{ {n}^{6 log_{a}(3) } } }}$

Step-by-step explanation:

Given,

$\frac{1}{2} log_{3}(m) + 3 log_{a}(n) = 1$

To express m in terms of n,

Further solving, we will get,

$= > \frac{1}{2} log_{3}(m) = 1 - 3 log_{a}(n)$

But, we know that,

• $log_{a}(a) = 1$

Therefore, we will get,

$= > \frac{1}{2} log_{3}(m) = log_{a}(a) - 3 log_{a}(n)$

Also, we know that,

• $y log(x) = log( {x}^{y} )$

Therefore, we will get,

$= > log_{3}( {m}^{ \frac{1}{2} } ) = log_{a}(a) - log_{3}( {n}^{3} )$

But, we know that,

• $log(x) - log(y) = log( \frac{x}{y} )$

Therefore, we will get,

$= > log_{3}( \sqrt{m} ) = log_{a}( \dfrac{a}{ {n}^{3} } )$

Now, we know that,

• $log_{x}(y) = \dfrac{ log_{b}(y) }{ log_{b}(x) }$

Therefore, we will get,

$= > \dfrac{ log_{a}( \sqrt{m} ) }{ log_{a}(3) } = log_{a}( \dfrac{a}{ {n}^{3} } ) \\ \\ = > log_{a}( \sqrt{m} ) = log_{a}( \dfrac{a}{ {n}^{ 3} } ) \times log_{a}(3) \\ \\ = > \sqrt{m} = {a}^{ log_{a}( \frac{a}{ {n}^{3} } ) log_{a}(3) } \\ \\ = > \sqrt{m} = { (\frac{a}{ {n}^{3} } )}^{ log_{a}(3) } \\ \\ = > \sqrt{m} = \frac{ {a}^{ log_{a}(3) } }{ {n}^{3 log_{a}(3) } } \\ \\ = > \sqrt{m} = \frac{3}{ {n}^{ log_{a}( {3}^{3} ) } } \\ \\ = > \sqrt{m} = \frac{3}{ {n}^{ log_{a}(27) } } \\ \\ = > \bold{ m = \dfrac{9}{ {n}^{6 log_{a}(3) } } }$

Answer 2

Answer:

change base 3 with a and u will get the answer