Q16. log₂[log₂{log₂(log₃81)}] = ?
(1) 1
(2) 0
(3) log3
(4) undefined
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Q16. log₂[log₂{log₂(log₃81)}] = ?(1) 1(2) 0(3) log3(4) undefined
Q16. log₂[log₂{log₂(log₃81)}] = ?
(1) 1
(2) 0
(3) log3
(4) undefined
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Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
We know that
[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ log_{x}( {x}^{y} ) = y}} [/tex]
Let's solve the problem now!!!
[tex]\large\underline{\sf{Given \: that- }}[/tex]
[tex]\rm :\longmapsto\: log_{2}( log_{2}( log_{2}( log_{3}(81) ) ) ) [/tex]
[tex]\rm \: = \: \: \: log_{2}( log_{2}( log_{2}( log_{3}( {3}^{4} ) ) ) ) [/tex]
[tex]\rm \: = \: \: \: log_{2}( log_{2}( log_{2}( 4 ) ) ) [/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \bf{ log_{x}( {x}^{y} ) = y}} [/tex]
[tex]\rm \: = \: \: \: log_{2}( log_{2}( log_{2}( {2}^{2} ) ) ) [/tex]
[tex]\rm \: = \: \: \: log_{2}( log_{2}(2 ) ) [/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \bf{ log_{x}( {x}^{y} ) = y}} [/tex]
[tex]\rm \: = \: \: \: log_{2}(1) [/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \bf{ log_{x}( {x}) = 1}} [/tex]
[tex]\rm \: = \: \: \:0[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: \bf{ log1= 0}} [/tex]
[tex]\bf :\implies\: log_{2}( log_{2}( log_{2}( log_{3}(81) ) ) ) = 0[/tex]
Hence,
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (2) \: is \: correct}}}[/tex]
Additional Information :-
[tex]\rm :\longmapsto\: log(xy) = log(x) + log(y) [/tex]
[tex]\rm :\longmapsto\: log \dfrac{x}{y} = log(x) - log(y) [/tex]
[tex]\rm :\longmapsto\: log( {x}^{y} ) = y log(x) [/tex]
[tex]\rm :\longmapsto\: log_{x}(x) = 1[/tex]
[tex]\rm :\longmapsto\: log_{ {x}^{m} }( {x}^{n} ) = \dfrac{m}{n} [/tex]
[tex]\rm :\longmapsto\: {e}^{logx} = x[/tex]
[tex]\rm :\longmapsto\: {e}^{ylogx} = {x}^{y} [/tex]
[tex]\rm :\longmapsto\: log(1) = 0[/tex]