Question

If log 81 base 12 =x,then log 3 base 6= A)x/x+4 B)x+4/x C)2x/x+4 D)x+4/2x

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

If log 81 base 12 =x , then log 3 base 6

A)x/x+4

B)x+4/x

C)2x/x+4

D)x+4/2x

EVALUATION

Here it is given that

$\displaystyle \sf{ log_{12}(81) = x}$

$\displaystyle \sf{ \implies \: log_{12}( {3}^{4} ) = x}$

$\displaystyle \sf{ \implies \:4 log_{12}(3 ) = x}$

$\displaystyle \sf{ \implies \: log_{12}(3 ) = \frac{x}{4} }$

$\displaystyle \sf{ \implies \: log_{3}(12 ) = \frac{4}{x} }$

$\displaystyle \sf{ \implies \: log_{3}(3 \times 4 ) = \frac{4}{x} }$

$\displaystyle \sf{ \implies \: log_{3}(3 ) + log_{3}(4) = \frac{4}{x} }$

$\displaystyle \sf{ \implies \: 1 + log_{3}(4) = \frac{4}{x} }$

$\displaystyle \sf{ \implies \: log_{3}(4) = \frac{4}{x} - 1}$

$\displaystyle \sf{ \implies \: log_{3}( {2}^{2} ) = \frac{4 - x}{x} }$

$\displaystyle \sf{ \implies \: 2 log_{3}( 2 ) = \frac{4 - x}{x} }$

$\displaystyle \sf{ \implies \: log_{3}( 2 ) = \frac{4 - x}{2x} }$

Now

$\displaystyle \sf{ log_{6}( 3) }$

$\displaystyle \sf{ = \frac{1}{ log_{3}( 6) } }$

$\displaystyle \sf{ = \frac{1}{ log_{3}( 3 \times 2) } }$

$\displaystyle \sf{ = \frac{1}{ log_{3}( 3 ) + log_{3}(2) } }$

$\displaystyle \sf{ = \frac{1}{ 1 + log_{3}(2) } }$

$\displaystyle \sf{ = \frac{1}{ 1 + \frac{4 - x}{2x} } }$

$\displaystyle \sf{ = \frac{1}{ \frac{2x + 4 - x}{2x} } }$

$\displaystyle \sf{ = \frac{1}{ \frac{x + 4 }{2x} } }$

$\displaystyle \sf{ = \frac{2x}{ x + 4 } }$

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{ C) \: \: \: \frac{2x}{ x + 4 } }$

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Answer 2

Answer:

2x/x+4

I hope I helped you have a good day :)