## How to solve this question? log²6 (base3)-log²2( base3)={log²x-2}log 12(base3) ​

Question

How to solve this question?

log²6 (base3)-log²2( base3)={log²x-2}log 12(base3)

in progress 0
4 weeks 2021-08-17T15:23:32+00:00 2 Answers 0 views 0

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:

2. 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