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)

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

Answers ( )

    0
    2021-08-17T15:24:48+00:00

    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:

    hope hope it helps please mark it as a brainliest answer

    0
    2021-08-17T15:24:55+00:00

    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

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