How to solve this question?
log²6 (base3)-log²2( base3)={log²x-2}log 12(base3)
Answers
Step-by-step explanation:
2^(2 + log 3 to the base 2)
So OK let’s start.
2^(2+log 3 to the base 2)
we can write it as
2^(log4 to the base 2 + log 3 to the base2)——————-{log 4 to the base 2 = 2}
=2^(log(4 *3) to the base 2)
=2^(log 12 to the base 2)
=12^(log 2 to the base 2)
=12^1
=12
Answer:
3, 6, 12 are in GP { because you can see the common ratio is 2 (constant)}
now, take log base 2,
log{3 base 2} , { log{6 base 2}, log{12 base 2} are in AP .
how? Let see
log{3 base 2} = first term
log{6 base 2} = log{ 3×2 base 2}
= log{3 base 2} + log{ 2 base 2}
= log{3 base 2} + 1 = second term
log{12 base 2} = log{3×2² base 2}
= log{3 base 2} + 2log{2 base 2}
= log{3 base 2} +2 = 3rd term
we see that,
log{3 base 2}, log{3base 2} +1 , log{3 base 2} + 2 ,are in AP
hence,
log{3 base 2}, log{6 base 2} , log{12 base 2} are in AP
so,
1/log{3 base 2} , 1/log{6 base 2} , 1/log{12 base 2} are in HP
we know, a/c to properties of logrithmn ,
log{a base b} = 1/log{b base a} use this here,
1/{1/log(2 base 3) }, 1/{1/log(2 base 6)}, 1/log(2 base 12)} are in HP
so,
log(2 base 3), log(2 base 6) , log(2 base 12) are HP.
hence,
1/log(2 base 3) , 1/log(2 base 6) , 1/log(2 base 12) are in AP
hence proved //
[ note :- inverse of AP is in HP and vice-versa ]
Step-by-step explanation:
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